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Math 140 Lecture 1 Intervals, absolute value, domains, polynomials INTERVAL NOTATION.

xι[1, 3] iff 1 < x < 3 solid dot includesxι(1, 3) iff 1 < x < 3 hollow dot excludesxι iff 1 < x (1, ∞)xι iff x <-1 or 1 < x(−∞, −1] ∪ [ 1, ∞)

ABSOLUTE VALUE.|3| = 3, |-3| = 3, |0| = 0.

Negating a negative makes it positive, -(-3) = 3. In general|x| = x if x > 0. |x| = -x if x < 0.

Rewrite without | |’s: = since π − 2 π − 2 π > 2 ⇒ π − 2 > 0 = 2 − π −( 2 − π) = π − 2

Rewrite without | |’s: for xι(3, 5). x − 3 + x − 5xι(3, 5) ˆ x > 3ˆ ˆ . x − 3 > 0 x − 3 = (x − 3)xι(3, 5) ˆ x < 5ˆ ˆx − 5 < 0 x − 5 = −(x − 5) = −x + 5τ x − 3 + x − 5 = (x − 3) + (−x + 5) = 2

DISTANCE. = the distance between a and b. a − bThe distance between 7 and 3 = = 4. 7 − 3The distance between 3 and 7 = = 4. 3 − 7 = − 4

NOTE. .x < 3 iff −3 < x < 3 iff x ∈ (− 3, 3)

.x > 3 iff x < −3 or 3 < x iff x ∈ (−∞ , −3) ∪ ( 3, ∞)Write as an interval: { x : x + 7 < 3}

x + 7 < 3 iff −3 < x + 7 < 3 iff −10 < x < −4Answer: = (-10, -4).{ x : x + 7 < 3}

Write as a union of two intervals: { x : x − 5 ≥ 2} x − 5 ≥ 2 iff x − 5 ≤ −2 or 2 ≤ x − 5

. iff x ≤ 3 or 7 ≤ x Answer: { x : x − 5 ≥ 2} = (−∞, 3] ∪ [ 7, ∞).DEFINITION. The domain of a function (real values, real

variables) is the set of numbers for which it is defined.

Where is not defined? Its domain is x = 2, 3.x−1(x−2)(x−3)

Where is not defined? Its domain is x > 2.x − 2Know what the following terms mean: polynomial,

degree, coefficients, quadratic function, expanded form,factored form mean. Know how to factor a polynomialand use the quadratic formula. (x + 1)2 = x2 + 2x + 1

is the factored form, (x + 1)2

is the expanded form. x2 + 2x + 1

Solving inequalitiesAs with equalities, you may add or subtract anything

from both sides. You may multiply or divide both sides bya positive number. If you multiply or divide by a negativenumber, you must change the direction of the inequality(multiplying 5>3 by -1 gives -5 < -3). If you don’t know ifa number is positive or negative, don’t multiply or divideby it.

(divide by 2)2x ≤ 4 ⇒ x ≤ 2 (divide by -2; change sign direction)−2x ≤ 4 ⇒ x ≥ −2

doesn’t imply (don’t know if x > 0 or x< 0)x < 1x x2 < 1

To solve problems like , use the key-number method:x < 1x

Rewrite with 0 on the right. Factor the f(x) on the left.f(x) < 0, f(x) > 0, f(x) ≤ 0, or f(x) ≥ 0.

Find key numbers x where f (x) = 0 or f (x) is undefined. On the key intervals before, between, and after key

numbers, f (x) is either > 0 or < 0. To find out which,evaluate f (x) at some point inside the interval. Youdon’t have to find the value, just the sign.

Use ( )’s with <, with >, around +5, and where f (x)is undefined. Use [ ]’s if f (x) is defined and the inequality is < or > .

Solve for x: . Write as a single interval.x < 1x

x < 1x

iff x − 1x < 0

iff x2−1x < 0

iff τ(x−1)(x+1)

x < 0 f(x) = (x−1)(x+1)x

Key numbers: x = -1, 0, 1.Key intervals: (-5, -1), (-1, 0), (0, 1), (1, 5).

f (-2)= −, f (-½)= +, f (½)= −, f (2)= +. We picked -2ι(-5,-1), -½ι(-1,0), ½ι(0,1), 2ι(1,5).

τx < 1/x iff f(x) < 0 iff Answer: xιιιι(−5−5−5−5, -1)υυυυ(0, 1)

Solve for x: . Put answer in interval notation. x2 + x − 6 ≤ 01st factor. f (x) = . x2 + x − 6 = (x − 2)(x + 3)f (x) = 0 iff x = 2, -3. These are the key numbers. The 3 key intervals: (-5,-3], [-3,2], [2,5). Picking -4, 0, 3: f (-4)>0, f (0)<0, f (3)>0. τ iff x2 + x − 6 ≤ 0Answer: xιιιι[-3, 2]

Write as a union of two intervals: . x−1x2−x−2

≥ 0 iff . x−1

x2−x−2≥ 0 x−1

(x+1)(x−2) ≥ 0Key numbers: -1, 1, 2. Undefined at -1, 2Key intervals: (-5,-1), (-1,1], [1,2), (2,5) f (-2)<0, f (0)>0, f (3/2)<0, f (3)>0τ Answer: xιιιι(-1, 1]υυυυ(2,

5555)

oo o

Math 140 Lecture 2 Rectangular coordinates(a,b) is the point on the plane whose x-coordinate is a and

y-coordinate is b.

a

b

a,b

x

y

2

1 5

fig. 2( )

WARNING. (a,b) can mean a point or an interval. DEFINITION. The graph of an equation is the set of all

points which satisfy the equation. The graph of y = 0 is the x-axis. The graph of x = 0 is the y-axis. Does (5, 6) lie on y = x2 +1?

(x, y) = (5,6) lies on y = x2 +1 iff 6 = (5)2

+1 iff 6 = 26 iffno.

DISTANCE FORMULA. The distance d between points (x1, y1)and (x2, y2) is d = (x1 − x2)2 + (y1 − y2)2

The distance between (1,3) and (2,5) is (1 − 2)2 + (3 − 5)2 = 12 + 22 = 5

Circles and completing the squareCIRCLE EQUATION. The circle with radius r and center (a,b)

has equation: . (x − a)2 + (y − b)2 = r2

Proof. (x, y) is on this circle iff the distance between (x, y) and (a, b) is r iff (x − a)2 + (y − b)2 = riff . (x − a)2 + (y − b)2 = r2

Find the equation for the circle of radius 3 around (1, 2): (x − 1)2 + (y − 2)2 = 32

FACTS. For positive a, . For all a, .a = ( a )2 a2 = aFind the center and radius of the circle with equation:

(x + 1)2 + y2 = 8 (x − (−1))2 + (y − 0)2 = ( 8 )2 = ( 2 ⋅ 4 )2 = (2 2 )2

τcenter = (-1,0), radius = 2 2COMPLETING THE SQUARE. To make a perfect square,x2 + ax

add . . The latter is a “perfect”(a2)2 x2 + ax + (a

2)2 = (x + a2)2

square.

Find the center and radius of the circle with equation:4x2 + 4y2 + 4y − 79 = 0

x2 + y2 + y − 794 = 0

x2 + y2 + y = 794

(x − 0)2 + (y2 + y + 14) = 79

4 + 14 = 80

4 = 20(x − 0)2 + (y + 1

2)2 = ( 20 )2 = ( 4 ⋅ 5 )2

(x − 0)2 + (y − (−12))2 = (2 5 )2

τcenter = (0, -½), radius = 2 5

DEFINTION. An x-intercept is the x-coordinate of a pointwhere the graph crosses the x-axis. A y-intercept is they-coordinate of a point where the graph crosses they-axis. In fig. 2, 2 is the y-intercept; 1, 5 are thex-intercepts.

The graph crosses the x-axis when the y-coordinate is 0.To find the x-intercepts, set y to 0 and solve for x.

To find the y-intercepts, set x to 0 and solve for y. Find the x and y-intercepts.

y − x2 + 4 = 0x-intercepts (set y = 0):

0 − x2 + 4 = 0, − x2 = −4, x2 = 4, x = −2, 2The x-intercepts are -2, 2.

y-intercepts (set x = 0): y − 02 + 4 = 0, y = −4

The y-intercept is -4. Find the x and y-intercepts.

y − x2 − 4 = 0x-intercepts:

x-intercepts: none0 − x2 − 4 = 0, − x2 = 4, x2 = −4,y-intercept: 4. Straight lines: slopes and equationsDEFINITION. The slope of a line is the ratio of the vertical

(height) change in y over a horizontal change in x.THEOREM.

The slope of the line through (x1, y1) and (x2, y2):m = y2−y1

x2−x1

The equation of the line through (x1, y1) with slope m: y − y1 = m(x − x1)

The equation of the line with slope m and y-intercept b: y = mx + b

The slope of a horizontal line is 0. The slope of a vertical line is undefined. Equation for the horizontal line through (a, b): y = b.Equation of the vertical line through (a, b): x = a.

Estimatetheslopes

x

y

Find the slope of the line through (2, 3) and (4, 3). 0LINE EQUATION FORMAT. In homework and tests, line

equations must be in the one of these four forms: y = mx+b, y = mx, y = b, x = a.

Find the line through (2, 5) and (4, 1). Check your answer. m = (5−1)/(2−4) = -2, y−5 = -2(x−2), y = -2x + 9 Eq. of line with slope -8, y-intercept -6. y = -8x −−−− 6

Math 140 Lecture 3 Reminder: Gateway Exam week from Thursday.Factoring and rootsTHEOREM. If , . But a > 0 x2 − a = (x − a )(x + a )

has no roots and can’t be factored any more.x2 + a

DIVISION LAW. If p(x)/d(x) has quotient q(x) and remainderr(x) then

p(x)d(x) = q(x) + r(x)

d(x) . Multiply by d(x) to get p(x) = d(x)q(x) + r(x).

d(x) divides into p(x) evenly iff the remainder is 0 iff iff d(x) is a factor of p(x).p(x) = d(x)q(x)

If d(x) is a factor of p(x), the other factor of p(x) is thequotient factor q(x). To get this quotient factor divide:p(x)/d(x).

Given p(x)/d(x), divide to get the quotient q(x) andremainder r(x). Write the answer in division law form:

.p(x) = d(x)q(x) + r(x)x3+1x−1 , x3+1

x−1 = x2 + x + 1 + 2x−1 , x3 + 1 = (x − 1)(x2 + x + 1) + 2

Check that the answer is correct for x = 0. For x = 0, we get 0+1 = (-1)(0+0+1)+2, 1 = 1. ,

X-INTERCEPT. a is a root or zero of p(x) iff p(a) = 0.

THEOREM. a is a root of p(x) iff is a factor of p(x). (x − a)

To find all roots of p(x), completely factor p(x).

Factor the polynomial and find all roots. Root: -2x + 2

Fully factored as is, no roots.x2 + 2

Roots: x2 − 2 = (x + 2 )(x − 2 ) − 2 , 2

One repeated factor. Root: 2x2 − 4x + 4 = (x − 2)2

given that -1 is a root. x3 + 5x2 + 8x + 4

τwe divide by (x+1).(x − a) = (x − (−1)) = (x + 1)

x3 + 5x2 + 8x + 4(x + 1) = x2 + 4x + 4 = (x + 2)(x + 2)

Roots: -1, -2. x3 + 5x2 + 8x + 4 = (x + 1)(x + 2)2

given that 1 is a root. x3 − x2 − 2x + 2

.x3 − x2 − 2x + 2(x − 1) = x2 − 2 = (x + 2 )(x − 2 )

x3 − x2 − 2x + 2 = (x − 1)(x + 2 )(x − 2 )Roots: . 1, − 2 , 2

. Factor out the coefficient of x2; find the2x2 + 2x − 2roots with the quadratic formula; factor.

2(x2 + x − 1)

Roots: x = −b± b2−4ac2a =

−1± 12−4(1)(−1)

2(1) = −1+ 52 , −1− 5

2

Factorization: 2(x − −1+ 52 )(x − −1− 5

2 )

Functions DEFINITION. For sets A and B, a function from A to B

assigns a value in B to each x in A. The domain of ff(x) is A; the range of f is the set of all possible values . f(x)

f (x) = x2 is a function from real numbers to realnumbers.domain = since x2 is defined for all numbers.(−∞, ∞)range = since x2 can never be negative. [0, ∞)

NOTATION. Sometimes, instead of writing f (x) = x2, we define a function by writing y = x2.

Thus y is the value of the function. Since it depends on x,y is the dependent variable. Since x ranges freely overthe domain, it is the independent variable.

A function may assign only one value to each x. Thus y = is not a function. ± xOf f and g, which are functions? (f isn’t, g is)

f g

Write each domain in interval notation.

y = 1 − x (−∞, ∞)

y = 11 − x (−∞, 1) ∪ ( 1, ∞)

y = 1 − x (−∞, 1]

f (x) = x2. Simplify to an expanded polynomial.

f (2) 4

f (x) + 2 x2 + 2

xf (x) x3

f (x3) x6

(f (x))3 x6

f (f (x)) x4

f(x) − f(a)x − a = x2 − a2

x − a = x + a

f(x + h) − f(x)h

= (x + h)2 − x2

h= 2x + h

Math 140 Lecture 4 Gateway exam. Gateway practice exams included with lecture notes.

DEFINITION. The graph of a function f is the set of allpoints (x, y) such that y = f (x). The height y is thefunction’s value f (x).

FACT. A curve is the graph of a function iff no verticalline intersects it more than once.

CWhich curve is the graph of a function?

CFind the domain and range of f, g, and h.Domain: [-1,3]Range: [1,4]Find the values. f (1) = 4

f (2) = 1For what value of x is f (x) = 1?

x = 2Compute [f (x)�f (1)]/[x�1] when x = 3.

-½

Domain: (-1,3]Range: {2}Find the values. g(1) = 2

g(5) = undefined For what values of x is g(x) = 2?

(-1,3]

Domain: [0, 2)X(2, 4]Range: [1, 3)X(4, 5]Find the values. h(2) = undefined

h(4) =1

DEFINITION. f increases on an interval if the value f (x) increases as xmoves from left to right in the interval. f decreases if f (x) decreases as x goes left to right. c is a turning point of f if f increases on one side of cand decreases on the other. f (a) = c is a maximum value iff c is ! all other values. f (a) = c is a minimum value iff c is � all other values.

CFor f and h graphed above, fill in the table. Write “none” if thereare none.

f h

turning points x = 1, 2 none

maximum value f(1) = 4 h(0) = 5

minimum value f(2) = 1 h(4) = 1

interval(s) of increase [-1, 1], [2, 3] none

interval(s) of decrease [1, 2] [0, 2), (2, 4]

CDo the same for the functions f and g below.

f g

domain [-2, 5] [0, 4]

range [-1, 4] [0, 3]

turning points x = 0, 2 x = 2

maximum value f(0) = 4 g(2) = 3

minimum value f(2) = -1 g(0) = g(4) = 0

interval(s) of increase [-2, 0], [2, 5] [0, 2]

interval(s) of decrease [0, 2] [2, 4]

CFill in the table. f(x) = x2 − 1

domain (-�,-1]X[1,�)

range [0,�)

turning points none

maximum value none

minimum value f(-1) = f(1) = 0

interval(s) of increase [1,�)

interval(s) of decrease (-�, -1]

CGraph.

f(x) =

1x if x < 0x if x ≥ 0

CGraph.

f(x) =

1 if x < 0x if 0 ≤ x ≤ 11 if 1 < x

1

1

o

CGateway problem. Complete the square of the formula. = ... = 2x2 − 3x + 5 2(x − 3

4)2 + 318

f g

oo

o

f

g

h

Math 140 Lecture 5 Exam 1, next week, Lectures 1-6

Basic graphsKnow these graphs. 1, x, x2, x3, 1/x.

1/x2, , |x|, .x 1 − x2

Translations and reflections THEOREM. Translating a graph f (x) or reflecting it across

an axis, changes the function as follows:

up 1 unit down 1 *left 1 *right 1reflect inx-axis

reflect iny-axis

f (x)+1 f (x)−1 f (x+1) f (x−1) -f (x) f (-x)* Horizontal changes are the opposite of what one would expect.

Given f(x), find the functions for the other graphs.

f(x)

1 2-1

-1

1

2right 2, up 1

right 2down 1

reflect in x-axis

reflect in y-axis

reflect in x and y-axis

f(-x)

-f(-x) -f(x)

f(x)-1

f(x-2)

f(x-2)+1

y-axis

x-axis3

VERTICAL MOVES

Given f (x) = |x|, graph f (x)+2, f (x)−2, -f (x). f (x) = |x|, f (x)+2= |x|+2, f (x)−2= |x|−2, -f (x) = -|x|.

The value f (x) = the height = the vertical position of apoint on the graph. Changing f (x) changes the verticalposition of the graph. Adding 2 raises the graph 2 units. Negating f (x) reflects the graph vertically across the x-axis.

Given f (x) = , graph f (x+2), f (x−2), f (-x).xf (x) = , f (x+2) = , f (x−2) = , f (-x).x x + 2 x − 2

HORIZONTAL MOVES

x = the x-axis position. Changing x, changes thehorizontal position of the coordinate system. Replacing xby x+2 shifts the coordinate system 2 units to the right.

Moving the coordinate system to the right equals movingthe graph to the left. Stated in terms of the graph ratherthan the coordinate system, changing x changes theposition of the graph in the opposite direction.

Replacing x by x+2 shifts the graph left 2 units. Replacing x by x−2 shifts the graph right 2 units. Replacing x by -x reflects the graph horizontally

across the y-axis.

. Describe the following shifts. f(x) = x

f (x) As abovef (x−1) = 1/(x−1) Shift right 1f (x+1) = 1/(x+1) Shift left 1f (-x) = 1/-x Reflect in y-axisf (-x+1) = 1/(1−x)Horizontal changes go in opposite the expected order. Start with f(-x+1)at the right and work backwards, right to left. At each step replace x withthe next larger term.f(x) →

← f(x + 1) →← f(−x + 1)

The graphing steps, executed left to right are:Shift left 1, reflect in y-axis

f (-(x+1)) = -1/(x+1)f(x) →

← f(−x) →← f(−(x + 1)).

Reflect in y-axis, shift left 1

f (x)+1 = (1/x)+1 Shift up 1-f (x) = -1/x Reflect in x-axis-f (-x+1) +1 = [-1/(-x+1)]+1 Start with in the middle. Work backwards for the horizontalf(−x + 1)and forwards for the vertical changes.f(x) →

← f(x + 1) →← f(−x + 1) → −f(−x + 1) → −f(−x + 1) + 1

Left 1, reflect across y-axis, reflect across x-axis, up 1.

Graph Parabola with roots 0, 1g(x) = x2 − x.Graph Shift left 1g(1 + x)Graph Shift left 1, reflect in y-axisg(1 − x)

Math 140 Lecture 6 Study Practice Exam 1 and the recommended exercises. Functions can be added and multiplied just like numbers.DEFINITION. For functions f and g, define f +g, f − g, f.g, f /g

by (f + g)(x) = f (x) +g(x), Note: (f+g)(x) is not (f+g).(x)(f − g)(x)= f (x)− g(x), The first is function application.(fg)(x) = f (x)g(x), The second is multiplication.(f /g)(x) = f (x)/g(x).

If f (x) = x −2, g(x) = 6, then (f+g)(x) = f (x)+g(x) = (x− 2) + (6) = x + 4, (f−g)(x) = f (x)−g(x) = (x − 2) − (6) = x − 8, (fg)(x) = f (x)g(x) = (x−2)(6) = 6x−12, (f /g)(x) = f (x)/g(x) = (x−2)/6 = .1

6x − 13

DEFINITION. For functions f and g, define fοg, thecomposition of f and g, by

(fοg)(x) = f (g(x)) Apply g to x. Get g(x). Apply f to g(x). Get f (g(x)). f is the outer function; g is the inner function.

Suppose f (x) = x −2 and g(x) = x2. (a) Find fοg and gοf.

(fοg)(x) = f(g(x)) = f(x2) = x2-2.(gοf)(x) = g(f(x)) = g(x-2) = (x-2)2 = x2-4x+4.

Note that f οg = gοf. For composition, order matters.(b) Find (fοg)(2).

Since (fοg)(x) = x2-2, (fοg)(2) = 22-2 = 4-2 = 2.(b’) Find (fοg)(2) and (gοf )(2) directly without (fοg)(x),

(gοf )(x).(fοg)(2) = f(g(2)) = f(22) = f(4) = 4-2 = 2.(gοf)(2) = g(f(2)) = g(2-2) = g(0) = 02 = 0.

If h(x) = c, then h(8) = c, h(y) = c, h(x2−1) = c, h(g(x)) = c.

For f(x) = 3x+4, g(x) = 5, find f οg and gοf.f(g(x)) = f(5) = 3.5+4 = 19.g(f(x)) = g(3x+4) = 5.

For f and g above, note that f(-3) = 1, f(-1) = 2, f(2) = 3, f(4) = 2, and g(-2) = -1, g(-1) = -2, g(1) = -1, g(2) = 2.

Find (fοg)(2) = f(g(2)) = f(2) = 3. (gοf)(2) = g(f(2)) = g(3) = undef. (fοf)(-1) = f(f(-1)) = f(2) = 3.

f(x) = x + 1x , f(f(x)) = (x + 1

x ) + 1/(x + 1x ) = x + 1

x + xx2+1

Write each function below as a composition f (g(x)) of two simpler functions,

an outer function f and an inner function g.

Find the inner function first. Write as a composition . (x2 + 2)6 f(g(x))

inner function (x2 + 2)6 g(x) = x2 + 2outer function does what remains f(x)to be done. τ

.f(x) = x6

check: . f(g(x)) = f (x2 + 2) = (x2 + 2)6

Write as a composition . 41x + 3 f (g(x))

inner function 4(1x ) + 3 g(x) = 1

x

outer function does what remains f (x)to be done. τ

. f (x) = 4x + 3check: . f (g(x)) = f (1

x ) = 4(1x ) + 3

Write as a composition. x + 1x + 1

g(x) = x + 1f (x) = x

τ .f (g(x)) = f (x + 1) = x + 1

Write as a composition.x4 + x2 + 1. x4 + x2 + 1 = (x2)2 + (x2) + 1

g(x) = x2

f (x) = x2 + x + 1τ .f (g(x)) = f(x2) = x4 + x2 + 1

Write as a composition of 2 functions.x /(1 + x )Write as a composition of 3 functions.1/(1 + x )

=h(f(g(x))), g(x) = x , f(x) = 1 + x, h(x) = 1/x

DEFINITION. id(x) = x is called the identity function.

id(x)=x

Hence id(5) = 5, id(y) = y, id(x2−1) = x2

−1, ... .THEOREM. For any function f (x), fοid = f and idοf = f.PROOF. ( f οid)(x) = f (id(x)) = f (x).

(idοf )(x) = id(f (x)) = f (x).0 is the identity for addition, since f + 0 = 0 + f = f . 1 is the identity for muliplication, f .1 = 1. f = f . id(x) is the identity for composition, since

fοid = idοf = f .

f(x)

g(x)

Math 140 Lecture 7 Inverse functionsDEFINITION. f

-1, the inverse of f , is the function, if any,such that, f ( f

-1(x)) = x when f -1(x) is defined and

f -1( f (x)) = x when f (x) is defined.

This says that f and f -1 undo each other: f -1 undoes what f (x) does and gives you back x.

f (x) = 2x, f -1(x) = ½x. Verify: f (f -1(x)) = x & f

-1(f (x)) = x g(x) = x +3, g-1(x) = x−3. h(x) = 2x +3, h

-1(x) = (x−3)/2 = . 12x − 3

2Verify that h

-1(h(x)) = x. To undo a sequence of operations, you must undo them in

the reverse order: the inverse of .g(f (x)) is f −1(g−1(x))Let y = f

-1(x). f (f -1(x)) = x, by definition of inverse. τf (y) = x, since y = f

-1(x). The converse is also true, thus THEOREM. y = f

-1(x) iff f (y) = x. To find f

-1(x) for complicated functions: Start with f (y) = x. Solve for y to get y = f

-1(x). , find . f (x) = x3 f −1(x)

f (y) = xτy3 = xτ .y = 3 x f −1(x) = 3 x

WARNING. f -1(x) and (f (x)) -1 are not the same.

f -1(x) = the inverse

(f (x)) -1 = the reciprocal = 1/f (x) .If f (x) = x3

f -1(x) = f

-1(0) = 03 x(f (x))

-1 = 1/x3 (f (0)) -1 = undefined.

= , find .f (x) x+1x−1 f −1(x)

f (y) = x, y+1y−1 = x, y + 1 = x(y − 1)

y + 1 = xy − x, y − xy = −x − 1y(1 − x) = −x − 1, y = −x−1

1−x = x+1x−1

τf −1(x) = x+1x−1

If f (x) = x+3 then f -1(x) = x −3.

If g(x) = x/2 then g -1(x) = 2x.

If h(x) = then h -1(x) = x2 for x > 0. x

Note how the graph of f is related to the graph of f -1.

f f-1

-1

-1g

g h

h

By the Theorem, y = f -1(x) iff f (y) = x. Thus the graph of

y = f -1(x) is the graph of f (y) = x which is just the graph

of f (x) = y with x and y interchanged. Interchanging xand y reflects the plane around the major diagonal y = x.Hence

THEOREM. The graph of y = f -1(x) is the reflection of thegraph of y = f (x) across the major diagonal y = x.

For each function, draw the three graphs y = f (x), y = x, y = f -1(x) on the same coordinate system.

f (x) = x3 f (x) = -x

3

DEFINITION. f is 1-1 (“one-to-one”) iff x = y implies f (x) = f (y).

f (x) = 3x is 1-1 x = y ˆ 3x = 3y f (x) = x

2 is not 1 = -1 but (-1) 2

= 12.

THEOREM. The following are equivalent: f has an inverse

f is 1-1

no horizontal line intersects its graph more than once.

Which of the following functions has an inverse?

THEOREM. The domain of f -1 is the range of f. The range of

f -1 is the domain of f .

Proof. The reflection around the major diagonal whichcarries the graph of f to the graph of f

-1 also carries thedomain of f to the range of f

-1 and the range of f to thedomain of f

-1.Stated in full, the inverse is the compositional inverse.

Compare it with the additive inverse and themultiplicative inverse.

For addition, 0 is the identity and the additive inverse of fis the negative -f since f + (-f ) = (-f ) + f = 0.

For multiplication, 1 is the identity and the multiplicative inverse of f is the reciprocal 1/f since:f . (1/f ) = (1/f ) . f = 1.

For composition, id is the identity where id(x) = x and for the inverse f

-1 of f, the corresponding equation is f ο f

-1 = f

-1ο f = id

Math 140 Lecture 8 DEFINITION. A quadratic function is a degree-2 polynomial

with a = 0. y = ax2 + bx + cThe graph is a parabola.

If a > 0, the horns point up. If a < 0, the horns point down. If |a| > 1, the parabola is narrower than y = x2. If |a| < 1, the parabola is wider than y = x2.

Find the roots by factoring or using the quadratic formula:

. No roots if .−b ± b2 − 4ac2a b2 − 4ac < 0

COMPLETING THE SQUARE THEOREM. Every quadraticfunction may be written in the form: y = a(x − x0)2 + y0where (x0, y0) is the vertex (nose) of the parabola.

Proof. Factor the a out of the ax2+bx part of ax2+bx+c .Complete the square. Anything which is added mustalso be subtracted to preserve equality.

EXAMPLES

Find the roots (they are the x-intercepts). Write in the form . y = a(x − x0)2 + y0Graph. On the graph list both coordinates of the vertex.

y = −12(x + 1)2

Roots: x = -1 vertex: (-1, 0)y = −1

2(x − (−1))2 + 0

(-1, 0)

y = 2x2 − 2x y = 2x2 − 2x = 2x(x − 1),

Roots: 0, 1. Factor out coeff. of x

2.y = 2(x2 − x) -1,;by2,ƒ(-1/2),square,ƒ (1/4)y = 2(x2 − x + 1

4) − 2 ⋅ 14

The added 1/4 is multiplied by 2, thus so is the subtracted 1/4.. vertex: (1/2, -1/2)y = 2(x − 1

2)2 + (−12 )

y = x2 + 2x − 1

Roots: x = −b± b2−4ac2a = −2± 4+4

2 = −1 ± 2 y = (x2 + 2x + 1) − 1 − 1

y = (x + 1)2 − 2. vertex: (-1, -2)y = (x − (−1))2 + (−2)

(-1,-2)

Word problemsDraw the picture. Indicate the variables in the picture. Write the given equations which relate the variables. Solve for the wanted quantities. List Given and Answer.

The perimeter of a rectangle is 10. Express the area A in terms ofthe width x.

Picture:

Given: 10 = 2x + 2yA = xy

Want A in x, need y in x: 2x+2y = 10

τx+y = 5 τy = 5 − xτA = xy = x(5 − x)

Answer: A = .x(5 − x)

The corner of a triangle lies on the line y = . Express the area A4 − xand perimeter P of the triangle in terms of the base x.

Given: y = 4 − xA = xy

2P = x + y + zz2 = x2 + y2

Want A in x, need y in x: Since , y = 4 − x A = xy/2 = x(4 − x)/2

Answer: A = x(4 − x)2

Want P in x, have y in x, need z in x: z = x2 + y2 = x2 + (4 − x)2

τP = x + y + z = x + (4 − x) + x2 + (16 − 8x + x2)Answer: .P = 4 + 2x2 − 8x + 16

The area of an isosceles triangle is 16. Express the height of thetriangle in terms of its width x.

Given: 12xh = 16

Want h in x. Answer: .h = 32x

(1/2,-1/2)

01

xy

10 = perimeterA = area

xy

y = 4-x

A = areaP = perimeter

z

h

x

16 = area

Math 140 Lecture 9 See inside text’s front cover for area and volume formulas.

Express the height of an equilateral triangle as afunction of a side x.

Given: h2 + (x2)2 = x2

Want h in x. 5 symbolsh2 = x2 − x2

4

h = 32 x

The height of a can (right circular cylinder) is three timesthe radius.

hπ2 r

Given: h = 3r, S = 2πrh(a) Express the curved surface area as a function of theradius. 4 symbolsS = 6πr2

(b) Express the radius as a function of the curvedsurface area. 5 symbolsr2 = S

6π

r = S6π

Three sides of a 500 square foot rectangle are fenced.Express the length of the fence as a function of side x.

Given: xy = 500 f = 2x + yWant f in x, need y in x.y = 500

xf = 2x + 500

x

Graphing polynomials A polynomial graph is smooth: no breaks, no sharp corners.For an expanded polynomial with axn the termaxn + ... + c

of highest degree: axn is the leading term, a is the leading coefficient, and n is the degree. The y-interceptis the constant term c.

, y = −3x4 + x2 − 5leading term = -3x4, leading coeff.= -3, degree= 4, constant term = -5.

, y = x − x3

leading term = -x3, leading coeff. = -1, degree = 3, constant term = 0.

y= x y= -x

y= xy= x

y= x 2 2

43 3y= -x

For large x (near +5), graph looks like the leading term axn.As x goes to 5, y goes to +5 if a > 0, to -5 if a < 0. Graphs of odd degree go to +5 in one direction, -5 inthe other, like y = x3, y = -x3. Graphs of even degree either go to +5 in bothdirections or to -5 in both directions, like y = x2, y = -x2.

Use the factored form to get the roots and their degrees(the degree of a root is the exponent of its factor). At roots of degree 1, the graph crosses x-axis like y = x or y = -x.At roots of odd degree n>1, the graph crosses the x-axis like y = x3

or y = -x3.At roots of even degree, the graph touches but doesn’t cross thex-axis, like y = x2 or y = -x2.

To get the leading coefficient and degree of a factored polynomial,replace each factor by its leading term.

To get the constant term (y-intercept), set x = 0.When graphing, find the x and y-intercepts and also calculate a key

value in each of the key intervals before, after, and between roots.

Graph y = (x+2)3(4 −x)(2x−1)2.-2 is a root of degree 3, 4 is a root of degree 1, ½ is a root ofdegree 2.lead term = (x)3(-x)(2x)2

= -4x6, degree = 6, lead coeff .= -4.Constant: setting x = 0, (2)3(4)(-1)2

= 32, y-intercept = 32.Values in the key intervals (-5,-2], [-2,½], [½,4], [4,5):f(-3) = -343, f(-1) = 45, f(2) = 1152, f(5) = -27783.

41/2-2

32

deg=1deg=2deg=3

100

-200

-300

1000

The dotted lines indicate the portion of the graph around f(2)= 1152which exceeds our coordinate limits.

hx

x/2 x/2

rh

S = curved surface area

x x

y

Area = 500f = length of fence

Math 140 Lecture 10

Rational functions and their graphsDEFINTION. A rational function is a ratio of two

polynomials. It is reduced if the top and bottom have nocommon factors.

Like polynomials, rational functions have smooth graphs.

CIn the graphs below, x = 0 (the y-axis) is the verticalasymptote, y = 0 (the x-axis) is the horizontal asymptote.

y=1/x

y= -1/x y= -1/x

y=1/x

y= -1/x

y=1/x 2

2 3

3

y=0

y=0

x=0

x=0

x=0y=0

x=0y=0

x=0

y=0

y=0

x=0vert

vert

hor

hor

For odd degree vertical asymptotes, one side goes to+�, the other to -�. See 1/x and -1/x.

For even degree vertical asymptotes, both sides go to+� or both go to -�. See 1/x2 and -1/x2.

DEFINITION. For rational functions, the leading term is thesimplified ratio of the leading terms of the top andbottom.

Recall, to get the leading term of a factored polynomial,replace each factor to its leading term and then simplify.

CRational functions:

Leading terms:

Hor. asymptote:

For a reduced rational function: x-intercepts (roots) occur where the top is 0.

If the root has degree n, the x-intercept looks like that of y = xn or y= -xn.

If the bottom is 0 at a, then x=a is a vertical asymptote.If the factor has degree n, the vertical asymptote looks like that of y=1/xn or y = �1/xn.

As x�, the graph resembles the graph of the leading term which is either a constant b or of the forma/xn or axn. ��� If a constant b, then y = b is a horizontal asymptote. ��� If it is a/xn, then y = 0 is a horizontal asymptote. ��� If it is axn, there is no horizontal asymptote.

Graph. On the graph mark the x and y-intercepts. Markthe vertical and horizontal asymptotes with theirequations ( y = a or x = a ).

a is a key number if f (a) is 0 or undefined. The keyintervals lie before, between and after the key numbers.Calculate a “key value” from each key interval.

Cy = 2x−62x3−8x2

reduce and factor: = = = 2(x−3)

2(x3−4x2)(x−3)

x3−4x2x−3

x2(x−4)y-intercept: nonex-intercept: 3 (deg 1)

vertical asymptotes: x = 0 (deg 2), x = 4 (deg 1)

lead term: 2x

2x3 = 1x2

horizontal asymptote: y = 0

key values: f (-1) = 4/5, f (1) = 2/3, f (7/2) = -4/49, f (5) = 2/25

C . y = 2x3−8x2

2x−6

reduce and factor: x2(x−4)x−3

y-intercept: 0 x-intercepts: 0 (deg 2),

4 (deg 1)

vert. asymp.: x = 3 (deg 1)

lead term: 2x3

2x= x2

hor. asymp.: none

key values: f (-1) = 5/4, f (1) = 3/2, f (7/2) = -49/4, f (5) = 25/2

0

x=3

4

v.a.

0

x=4

4y=0hor asym

vert

x=0vert asym

asym

x1−x

x2+13x2

(x−1)(5x+2)(x−3)2(1−2x)3

(1−2x)3

2x−6

−1 13

−58x3 −4x2

y = −1 y = 13 y = 0 none

Math 140 Lecture 11 Exam 2 next weekExponential functions DEFINITION. An exponential function is of the form y = bx

with the base b > 0. b0 = 1, b1 = b, b2 = b ⋅ b, ...b−n = 1/bn

, the nth root of b b1/n = n bbp/q = bp(1/q) = (bp)1/q = (b1/q)p

EXPONENT RULES Examples (bn)m = bnm (52)3 = (52)(52)(52) = 56

bnbm = bn+m 5253 = (5 ⋅ 5)(5 ⋅ 5 ⋅ 5) = 55

, bn

bm = bn−m 53

57 = 5−4 = 154

57

53 = 54

(ab)n = anbn 2535 = 65

(ab)n = an

bn (23)5 = 25

35

Simplify to an integer or a single exponent bn.

21/2

23/2

= 21/2+3/2

= (21/2)4/2 = 24/4 = 21 = 2

(2 3 ) 3 = 2 3 3 = 23 = 8

32+ 5

32− 5

= 32+ 5 +2− 5 = 34 = 81

81−π

81+π = 8(1−π)−(1+π) = 81−π−1−π = 8−2π

PROPERTY. If b = 1, . Hence is 1-1. bx = by ⇔ x = y bx

Gateway problems with the variable in the exponent: putboth sides above a common base then equate exponents.

27x = 9 (33)x = 32

τ τ3x = 2, τx = 2/3 33x = 32

8z+1 = 32/ 2 8z+1 = 252−1/2

(23)z+1 = 25−12

23z+3 = 292

3z + 3 = 92

6z + 6 = 9 6z = 3 z = 1/2

, τ z8 = 32/ 2 z8 = 25/212 = 25−1

2 = 292 z = 2

916

Note, this is different, the variable is in the base ratherthan the exponent. Raise both sides to the 1/8 power.

means a is approximately equal to b.a ≈ bFact: , i.e., 210 is approximately equal to 103. 210 ≈ 103

Estimate 250 and 220. 250 = (210)5 ≈ (103)5 = 1015 220 = (210)2 ≈ (103)2 = 106

The graph of y = 1x is the horizontal line y = 1. Otherwise, the graph of y = bx

has y-intercept 1 but no x-intercept, it goes to 5 in one direction, it has the horizontal asymptote y = 0 in the other.

For b > 1, the graph of bx is like the graph of 2x as below. For 0 < b < 1, the graph is like (½)x.

y=2x

xy=(1/2)

1

y=0 hor asymp

Graph Reflect 2x across y-axis. y = 2−x

Graph Shift right 1 unit, down 1 unit. y = 2x−1 − 1

Graph Shift left 1, reflect across y-axis.y = 21−x

The function ex DEFINITION. ex is the unique exponential function bx whose

the tangent at (0, 1) has slope 1.Fact: Again, ≈ means approximately equal.e ≈ 2.7

Thus 2 < e < 3. Hence the graph of ex lies between thegraphs of 2x and 3x.

Similarly, e-x lies between 2-x and 3-x.

y = e x

-x y= e

1

y=0 h.a.

slope = 1

x-intercept noney-intercept 1hor. asym. y = 0 domain (-5, 5) range (0,5) True or false? RECALL: e ≈ 2.7

false since e < 1 1 < e ⇒ 1 < e ⇒ 1 < etrue since e2 < 9 e < 3 ⇒ e2 < 32 ⇒ e2 < 9

Math 140 Lecture 12 Exam 2 covers Lectures 7 -12. Study the recommended exercises.Review area, circumference, volume formulas - inside front cover.RECALL. The graphs of .ex and e−x

y = e x

-x y= e

1

y=0 h.a.

slope=1

x-intercept none y-intercept 1hor. asym. y = 0. vert. asym. nonedomain (-5,5) range (0,5), Graph . y = ex−1 − 1

Give the domain, range, intercepts and asymptotes.

1

-1 y=-1h.a.

x-intercept x = 1y-intercept y =

1e − 1 ≈ 1

2.7 − 1 ≈ −.63hor. asym. y = -1domain (-5,5) range (-1,5).

LogarithmsAssume b > 0, b = 1. Thus bx is 1-1 and it has an inverse.DEFINITION. , the log of x to the base b, is thelogb(x)

inverse of the exponential function bx. ln(x), the natural logarithm, = = the inverse of ex.loge(x)Note, “ln” is “el-n” not “one-n” or “eye-n”.

Inverses act in opposite directions and inverses cancel. τ iff . y = ln(x) iff .y = logb(x) by = x ey = x

ln(ex) = x. logb(bx) = x eln(x)

= x.b logb(x) = x

If we have ln(bx) instead of ln(ex), then ln and bx don’tcompletely cancel and ln(ex) = x becomes:

. The exponent comes down to the outside.ln(bx) = x ln bFACT. e0 = 1 ˆ 0 = ln(1). e1 = e ˆ 1 = ln(e)Simplify to a rational.

.log55 5 log55 5 = log55151/2 = log553/2 = 32

.log218 log2

18 = log2

123 = log22−3 = −3

.log82 log82 = log83 8 = log881/3 = 1/3

Write in log form. Either (method 1: easy problems) use thedefinition of log or ( method 2: hard problems) take the appropriatelog of both sides and cancel inverses.

23 = 8Method 1. Since 2x and log2(x) are inverse,

3 = log2(8) 52x−1 = 6

Method 2. Take the log of both sides. log5(52x−1) = log5(6) so 2x − 1 = log5(6)

Solve for x, write the answer using logarithms. 5x2 = 4

x2 = log5(4)Leave answers in exact form, no decimals.x = ± log5(4)

e3t+1 = 8 3t + 1 = ln(8) 3t = ln(8) − 1

t = 13(ln(8) − 1)

Use natural logarithms and method 2.32x = 5x+1

ln(32x) = ln(5x+1) 2x ln 3 = (x + 1)ln 5

x(2 ln 3 − ln 5) = ln 5See also Practice Exam 2, prob. 11b.x = ln 5/(2 ln 3 − ln 5)

Since they are inverses, the graph of is thelogb(x)reflection of bx across the major diagonal and ln(x) isthe reflection of ex.

For the graph of ln(x)x-intercept 1 y-intercept nonevert. asym. x = 0. hor. asym. nonedomain (0,5) range (-5,5).

Graph . y = ln(x + 1) + 1x=-1

vert asym

1

-..63

y-intercept: ln(0+1)+1 = ln1 +1 = 0+1 = 1x-intercept: ln(x+1)+1 = 0 iff ln(x+1) = -1 iff x+1 = e-1 iff

x = 1/e − 1 ≈ 1/(2.7)−1 ≈ -.63vert. asym: x = -1 hor. asym: nonedomain: (-1, 5) range: (-5,5).

Graph . Left 1, refl. across y, refl. across x, up 1.y = 1 − ln(1 − x)

y=ex

1 y=ln(x)

1

x=0 v.a.

Math 140 Lecture 13 Recall: Since and are inverses of each other:logbx bx

= y iff x = logbx by

logbbx = xb logbx = x

Most properties of logarithms are the same as forexponentiation but with

the symbols 0 + − n( _ )exchanged with 1 ξ ; ( _ )

n

Assume b > 0, b = 1. On the log side, assume x, y > 0.LOG PROPERTIES EXPONENT PROPERTIES

logb1 = 0 b0 = 1logbb = 1 b1 = blogbxy = logbx + logby bx+y = bxby

logbx/y = logbx − logby bx−y = bx/by

logbxn = n logbx bnx = (bx)n

Proofs. iff logbxy = logbx + logby

iff b logbxy = b logbx+log by

iff xy = blogbxb logby

iff true. xy = xyLikewise for . logb

xy = logbx − logby

iff logbxn = n logbx iff b logbxn = bn logbx

iff xn = b(logbx)n

iff xn = [b logbx] n

iff true. xn = xn

When b = e, logb(b) = 1 becomes: ln(e) = 1 Note: . The lastlogb(x ⋅ y) = logbx + logby ≠ logb(x + y)

term cannot be broken into simpler pieces.Simplify. ln e2 − ln e4 + ln 1 − ln e

= 2 − 4 + 0 − 1 = −3

e ln 2 − e ln 4 + e ln 1 − e ln e

= 2 − 4 + 1 − e = −1 − eCombine into a single logarithm. 2 log10x + log10y

= log10x2 + log10y = log10x2y

log2x − 4 log2y = log2x − log2y4 = log2( x

y4 )

ln(x2 − y2) − ln(y2 + 1) = ln x2−y2

y2+1

Write as a sum / difference / multiple of the simplestpossible logarithms.

logb 42y

y3−a

= logb

2yy3−a

14

= 14 logb

2yy3−a

= 14 [logb(2y) − logb(y3 − a)]

= 14 [logb2 + logby − logb(y3 − a)]

ln 1a4 + b4

= ln(a4 + b4)−1/2 = −12 ln(a4 + b4)

SOLVING FOR x.Put terms involving x on the left, the rest on the right.Combine into a single logarithm. Exponentiate both sides to the base of the logarithm. Solve. Delete invalid solutions that give undefinedlogarithms.

log24x − log23 = log2(x + 2) log24x − log2(x + 2) = log23

log24x

x+2 = log23

2log24x

x+2 = 2log23

4xx+2 = 3, 4x = 3x + 6, x = 6Validity: for x = 6, both arelog24x and log2(x + 2)defined, thus the solution is valid.

ln(x + 1) = 2 − ln(x − 1) ln(x + 1) + ln(x − 1) = 2

ln [(x + 1)(x − 1)] = 2 ln(x2 − 1) = 2

e ln(x2−1) = e2

x2 − 1 = e2

x2 = 1 + e2

x = ± 1 + e2

Validity: is not valid since x = − 1 + e2

x negative ˆ x−1 is negative ˆ ln(x−1) is undefined.The only valid solution is .x = 1 + e2

CHANGE OF BASE FORMULA. .logax = logbxlogba

Proof: , now dividelogbx = logb(a logax) = loga(x)logbaboth sides by log ba.

Express in terms of log10: logb2 logb2 =

log102log10b

Express in terms of natural logarithms ln: log5tlog5t = ln t

ln 5

Math 140 Lecture 14 Exponential growthA bacteria colony starts with 10 bugs. Each bug splits into two

bugs every hour. How many bugs are there after t hours?Number of hours Number of bugs

0 101 10.22 (10.2).2 =

10.22

3 (10.Ν22).2 = 10.23

... ...t hours 10.2t

Exponential functions measure the size of a growing population, theamount of money in a compound interest account, the number ofatoms left after radioactive decay, etc. N(t) = the amount at time t.

BASE-e FORM LEMMA. Every exponential function can bewritten in the form

. N(t) = N0ekt

N0 = N(0) is the initial amount. k, the coefficient of t, is the growth constant. If k > 0, N(t) measures exponential growth. If k < 0, N(t) measures exponential decay.

FACT: Every a > 0 is a power of e: . a = e ln a

Example: 2 = .e ln 2

Write the bug population in base-e form.N(t) = 10 ⋅ 2 t

N(t) = 10 ⋅ 2 t = 10 ⋅ (e ln 2)t = 10 ⋅ e [ln(2)] t

Hence, in base-e form, . N(t) = 10e [ln 2] t

Thus N0 = 10, the initial amount and k = ln(2), the natural log of the initial base. You put $4,000 in an account at 5% interest

compounded annually. Write the amount N(t) of moneyin the account after t years in base-e form.Number of years Amount N(t) in account

0 4,0001 4000+(.05)4000 =

4000.(1.05)2 4000.(1.05)+(.05)4000.(1.05) =

4000..(1.05).(1+.05) =4000.(1.05)2

3 4000.(1.05)3

... ...t years 4000.(1.05)t

Hence N(t) = 4000(1.05) t

= 4000(e [ln(1.05)] )t = 4000e [ln(1.05)] t

τ . N(t) = 4000e [ln 1.05] t

τN0 = 4000 and k = ln(1.05). In each problem, first write N(t) in base-e form, then solve.A bacteria colony starts with 103 bugs. Four hours later

it has 5ξ103 bugs.

(a) Find the growth constant k. (b) What is the population two hours after the start? (c) How long will it take for the population to triple?

First find N(t) = the number of bugs after t hours. Given:

N0 = 103 N(4) = 5ξ103 N(t) = N0ekt

τN(t) = 103ekt

103ek4 = N(4) = 5 ⋅ 103

e4k = 5 � may omit this alternate step ln(e4k) = ln 5

4k = ln 5 k = ln 5

4

� base-e form N(t) = 103eln54 t

(a) The growth constant is k = ln 54

(b) After two hours the population is N(2) = 103e

ln54 2 = 103e

ln52

(c) Let t be the time when the population has tripled. τ N(t) = 3 × the initial amount N0τ N(t) = 3 ⋅ 103

τ 103eln 54 t = 3 ⋅ 103

τ eln54 t = 3

τ ln 54 t = ln 3

τ t = 4 ln 3ln 5

The population triples after hours.4 ln 3ln 5

Initially a sample has 8 lbs of a radioactive substancewith a half-life of 5 days, i.e., half decays after 5 days.How many lbs remain after 3 days?

First find N(t) = the number of lbs left after t days.Given:

N0 = 8N(5) = 4

N(t) = N0ekt

τ N(t) = 8ekt

8ek5 = N(5) = 4 e5k = 1

2 5k = ln 1

2

k = 15 ln(1/2)

� base e form N(t) = 8e15 ln(1/2)t

N(3) = 8e15 ln(1/2)3 = 8e

35 ln 1/2

τ lbs remain after 3 days. 8e35 ln 1/2

This is a decay since ln(1/2) = ln(2-1) = -ln(2) is negative.

Math 140 Lecture 15 Angles, arcs, and radiansRECALL. A circle of radius r has circumference 2Sr. S�≈ 3.14. Unit circle circumference = 2Sr = 2S1 = 2S.

T and Z are the Greek letters “theta” and “omega”.

DEFINITION. Suppose the vertex of an angle is at the centerof a circle of radius r. Let s be the length of the arc theangle intercepts on the circle. Then

T�= s/r is the radian measure of the angle. For unit circles, the radius r = 1 and radian measure equals arc length: T = s.

CRadian and degree measures on the unit circle .

S��

S��

� �� ��� �S

��

��

�S�� ���

R

R

RR R

R

�

��� S

Clockwise angles "are negative.

CONVERSION FORMULAS. 180o =�S radians. Thus 1o =�S/180 radians; 1 radian = 180/S� degrees.

CConvert 100o to radians. ≈ 1.74 radians 100o = 100 π

180 radians= 5π9 radians

CConvert S/6 radians to degrees. . π

6 radians= π6

180π

o = 30o

When T is in radians, we can solve T = s/r for arclength: s = Tr

CFind the length of a 30o arc on a circle of radius 12inches. First convert to radians. By the above 30o

=�S/6 radians.W . s = θr = π

612 = 2π inches

CFind the radian measure of an angle which intercepts a5 inch arc on a circle of radius 12 inches.

radians θ = 512

Speed DEFINITION. If an object travels a distance d in time t, its

linear speed is d/t. If an object rotates through an angle T in time t, itsrotational speed is Z = T/t.

THEOREM. If a point rotates around a circle of radius rwith rotational speed Z� then its linear speed is Zr.

Proof. If a point rotates through an angle T on a circle ofradius r in time t, then Z�=�T/t and the distance d ittravels = the length of the arc it traces = Tr . �Wits linearspeed = d/t = Tr/t = (T/t)r = Zr .

CA point revolves around a circle of radius 3 feet at 10revolutions per minute.

�D� What is its rotational speed (in radians)? 1 revolution = 2S radians, WZ = 10 revs/min = 10�2S radians/min. = 20S radians/min.

�E� What is its linear speed? Its linear speed = Z r = (20S rad/min)� (3 feet) = 60S feet/min.

Radians are dimensionless; you may delete dimensionless units.

Trigonometric functions The unit circle has radius one and center (0, 0). There are

four quadrants I, II, III, IV as pictured. An angle is in standard position if its vertex is the origin

(0,0) and its initial side is the positive x-axis. The otherside of the angle is the terminal side. The terminal sideintersects the unit circle at a point P

T.

P

TI

II

III IV

x(0,0)

T

DEFINITION. Given a standard-position angle of radianmeasure T, let (x, y) be the coordinates of P

T.

The six trigonometric functions of �T are:

sin T = y sinecsc T = 1/ sin T cosecant

cos T = x cosinesec T = 1/cos T secant

tan T = sin T / cos T tangent cot T = cos T / sin T cotangent

CDraw an angle of S/2 radians in standard position. Findthe six trigonometric functions. sin(S/2) = 1 csc(S/2) = 1cos(S/2) = 0 sec(S/2) = undef tan(S/2) = undef cot(S/2) = 0

CA point (x, y) on the unit circle is in the secondquadrant and y = . Find the six trig functions for T.3

4

y = . First, find x.34

(x, y) on the unit circle î x

2 + y

2 = 1.

W .x2 = 1 − y2 = 1 − (34)2 = 1 − 9

16 = 716

x= . (x,y) in the second quadrant î x is negative.± 7 /4Wx = . − 7 /4sin T = 3/4 csc T = 4/3 cos T�= sec T = − 7 /4 −4/ 7tan T = cot T = −3/ 7 − 7 /3

1T

y = sinT

x=cosT

P = (x,y)T

3/4(x,y)

x

1T

Math 140 Lecture 16Fri. = last day to withdraw. Keller 419A secretary will sign for me.FACTS. Know the sin and cos (and hence tan, cot, sec, csc) of:

0, π/6, π/4, π/3, π/2.

0

π/6=30

π/4=45

π/3=60π/2=90ο

ο

ο

ο

ο

1

(0,1)

(1,0)

λ

λ λ

( 3/2,1/2)

(1/ 2, 1/ 2)

(1/ 2, 3/ 2) λ

sin(0) = 0 cos(0) = 1 tan(0) = 0sin(π/6) = 1/2 cos(π/6) = tan(π/6) = 1/λ33 /2sin(π/4) = cos(π/4) = * tan(π/4) = 11/ 2 1/ 2sin(π/3) = λ3/2 cos(π/3) = 1/2 tan(π/3) = λ3sin(π/2) = 1 cos(π/2) = 0 tan(π/2) = undef. * is also ok. 2 /2PROOFS. All points are on the unit circle.0 = 0o: The point for 0 = the rightmost point = (1,0). π/2 = 90o: The point for π/2 = the top point = (0,1).For the other angles, we have a right triangle whose

hypotenuse is a radus of length 1.π/6 = 30o: This is a 30o− 60o right triangle. Thus the small

leg is ½ since it is half the hypotenuse which is 1. Let xbe the third side.

11/2

x30

30o 60

60

o

o o1x

Then x2 + (1/2)2 = 12

x2 + 1/4 = 1,

x2 = 3/4

x = λ3/2 τ sin(π/6) = sin(30o) = 1/2 and cos(π/6)= 3 /2

π/3 = 60o: This is the 30o triangle again but with the 30o

and 60o angles swapped. Swapping sides 1/2 and 3 /2gives: sin(π/3) = sin(60o) = and cos(π/3) = 1/2. 3 /2

π/4 = 45o: If one angle of a right triangle is 45o, so is theother acute angle and the triangle is isosceles.

Let x be the side length.

1

xx

Then x2 + x2 = 12

2x2 = 1 x2 = 1/2, x = . 1/ 2τsin(π/4) = sin(45o) = = cos(π/4)1/ 2

DEFINITION. For any angle in standard position (vertex = theorigin, initial side = the positive x-axis), its reference angle isthe acute positive angle (thus in [0, π/2]) between thex-axis (positive or negative, whichever is nearest) and theangle’s terminal side.

ref

ref ref

THEOREM. The sin and cos of θ equals the sin and cos ofits reference angle except for the sign which isdetermined by θ’s quadrant.

Rewrite in terms of reference angles, then evaluate. cos(120o) = -cos(60o) = -1/2 quadrant IIsin(120o) = sin(60o) = quadrant II3 /2cos(3π/4) = -cos(π/4) = quadrant II−1/ 2sin(3π/4) = sin(π/4) = quadrant II1/ 2cos(-3π/4) = -cos(π/4) = quadrant III−1/ 2sin(-3π/4) = -sin(π/4) = quadrant III−1/ 2

List three angles (in radian measure) whose cos is ½. π/3, -π/3, 5π/3

NOTATION. sin θ means sin( θ); sin2 θ means (sin θ)2.

THEOREM. Since sin θ and cos θ are the legs of a righttriangle of hypotenuse 1,

sin2 θ + cos2

θ = 1 sin θ = -2/3 and π < θ < 3π/2. Find cos θ and tan θ.

(-2/3)2 + cos2

θ = 1 cos2

θ = 1 − 4/9 = 5/9 cos θ = ± 5 /3π < θ < 3π/2 is in quadrant III, thus cos θ < 0 τ

cos θ = − 5 /3tan θ = sin θ/cos θ = 2/ 5

sec θ = -3 and sin θ < 0. Find tan θ. First find sin, cos.cos θ = -1/3 sin

2 θ + (-1/3)2 = 1

sin 2

θ = 1−1/9 = 8/9 sin θ = . Since sin θ < 0, ± 8 /3sin θ = = . τtan θ = . − 8 /3 −2 2 /3 2 2

Simplify (cos θ + 1)(tan θ + sin θ) = cos θ tan θ+cos θ sin θ+tan θ+sin θ (tanθ = sinθ/cosθ)= sin θ + cos θ sin θ + tan θ + sin θ = 2sin θ + cos θ sin θ + tan θ.

Note y2 +3y +2 factors into (y +1)(y +2).

Factor: csc2 θ + 3csc θ + 2 = (csc θ+1)(csc θ+2).Note x2−y2 = (x−y)(x+y). Factor: sin2 B − cos2 B = (sin B−cos B)(sin B+cos B).

Math 140 Lecture 1 7 Exam 3 next weekRECALL. sin2

T+ cos2 T�= 1

tan T = sin T/cos T cot T�= cos T/sin T

sec T = 1/cos T csc T = 1/sin T

Simplify.

C cotθ secθ cosθ = cosθsinθ

1cosθ cosθ = cotθ

C

sinx−1sinx + 1

sinx+1sinx − 1

= sinx − 1 + sinxsinx + 1 − sinx

= 2 sinx − 11

= 2sinx − 1

DEFINITION. An identity is an equation which is alwaystrue.

Cx2 +y2

= 1, y = 2x aren’t identities: they aren’t always true.sin2

T+ cos2 T= 1, xy = yx are identities, i.e., always true.

To prove an identity either ��� start from one side (usually

the more complicated side) and work your way to the otherside or ��� show that the equality is equivalent to a truth.The text uses ��� we’ll give examples of ���.

CProve cot2θ = csc2θ − 1

iff cos2θsin2θ

= 1

sin2θ− 1 iff cos2θ = 1 − sin2θ

iff iff true.sin2θ + cos2θ = 1

CProve cosθcscθ = sinθ

secθ

(by cross multiplying)iff cosθ secθ = cscθ sinθ

true. iff cosθ 1cosθ = 1

sinθ sinθ iff 1 = 1 iff

Right triangles

T T

1y

ac

bx

The trig functions were defined on the unit circle wherethe hypotenuse of the right triangle had length 1. For theangle pictured, sin T�= y. Any other right triangle withan angle T is similar to this unit-circle triangle. Hencethe ratios of the corresponding sides are equal. Thus sin T = y = y/1 = a/c = side opposite/hypotenuse.

THEOREM. For any right triangle with an acute angle T:sin T = opposite/hypotenuse* cos T = adjacent/hypotenuse tan T = opposite/adjacent

* More precisely, “length of the side opposite/length of thehypotenuse”. We’ll abbreviate this to just: opp/hyp.

In a right triangle, the two complementary small anglesadd up to 90o. If one is T, the other is 90o �T or S/2 � T.

THEOREM. sin(π

2 − θ) = cosθ sin(900 − θ) = cosθcos(π

2 − θ) = sinθ cos(900 − θ) = sinθ

Proof. In the picture below,

T

���Tc

a

b

sin T and are both a/c. cos(900 − θ)cos T and are both b/c. sin(900 − θ)

6

5T

x-1

9x2

B��� T

Fig. 1 Fig. 2

CIn Fig. 1, find sin, cos and tan for T and . 900 − θFirst find the third side. The hypotenuse = 52 + 62 = 61sin T = 6/ 61cos T = 5/ 61tan T = 6/5 sin(90o − θ) = 5/ 61

cos(90o − θ) = 6/ 61 tan(90o − θ) = 5/6

CIn Fig. 2, find sin B, cos B and tan B.

Let y be the third side. Then y2 + (x − 1)2 = (9x2)2

y2 = 81x4 − (x − 1)2 = 81x4 − (x2 − 2x + 1) y2 = 81x4 − x2 + 2x− 1

y = 81x4 − x2 + 2x − 1sin B = y/9x2 = 81x4 − x2 + 2x − 1 /9x2

cos B = (x − 1)/(9x2)tan B = y/(x − 1) = 81x4 − x2 + 2x− 1 /(x − 1)

Ctan T = 2/3, find sin T and cos T� where T is acute.

Since tan = opposite/adjacent = 2/3, draw a right trianglewith opposite = 2 and adjacent = 3 (the drawing need not be

exact). Then find the third side.

�

�

y

T

Then y = 32 + 22 = 13sin T = 2/ 13cos T = 3/ 13

Math 140 Lecture 18 Exam 3 covers lectures 12 -18. Study the recommended exercises.

DEFINITION. For any function f: f is even iff f (-x) = f (x). f is odd iff f (-x) = - f (x).

Graphically, the left half (the half plane left of the y-axis) of aneven function is of the right half across the y-axis. Theleft half of an odd function is the negative of thisreflection.

f(x) = x2 is even since f (-x) = (-x) 2 = x

2 = f (x). g(x) = x3 is odd since g(-x) = (-x)

3 = -x

3 = -g(x).

xx

2 3

even odd

Instead of thinking of sin θ as a function of an angle θ, wecan think of it as a function sin(t) of a real variable t.

MINUS THEOREM. sin(-t) = -sin(t), cos(-t) = cos(t), tan(-t) =-tan(t). Thus cos(t) is even; sin(t) and tan(t) are odd functions.

Proof.

t1

-t

sin(t)

sin(-t)=-sin(t)cos(-t)=cos(t)

tan(-t) = sin(-t)/cos(-t) = -sin(t)/cos(t) = -tan(t).

DEFINITION. A function f is periodic with period p iff f(x + p) = f(x)

for all x. p must be the smallest such positive number.

2π THEOREM. sin(t +2π) = sin(t), cos(t +2π) = cos(t), forintegers k, sin(t+2kπ) = sin(t), cos(t+2kπ) = cos(t).

Proof.

tt +π

(cos t, sin t)=(cos(t+2 ),sin(t+2 ))

(-cos t, -sin t) = (cos(t+ ), sin(t+ ))

t+2π

π π

π π

π THEOREM. sin(t +π) = -sin(t), cos(t+π) = -cos(t), tan(t +π)= tan(t).

Proof. See the picture for sin and cos. tan(t+π) =sin(t+π)/cos(t+π) = -sin(t)/-cos(t) = sin(t)/cos(t) = tan(t).

Thus sin and cos have period 2π, but tan has period π.Simplify: .cos(3π −x)

cos(3π −x) = cos(2π + π −x) = cos(π −x)= −cos(−x) = −cos(x).

Previously we found reference angles in [0, π/2]geometrically. Now we can find them algebraically.

Rewrite in terms of the nearest multiple of π. Throw outthis multiple of π using the 2π or π Theorem. Handleminus signs using the Minus Theorem.

Rewrite with a reference angle in [0,π/2], then evaluate.sin(5π/6). 5π/6 is near 6π/6 = π.

sin(5π/6) = sin(π−π/6) = -sin(-π/6) = sin(π/6) = 1/2.

cos(4 π/3). 4π/3 is near 3π/3 = π.cos(4π/3)=cos(3π/3+π/3)=cos(π+π/3) = -cos(π/3) = -1/2.

tan(-7 π/4). tan(-7π/4) = -tan(7 π/4). 7π/4 is near 8π/4-tan(7π/4) = -tan(8π/4−π/4) = -tan(2π−π4) = -tan(-π/4)= tan(π/4) = sin(π/4)/cos(π/4) = 1.

sin(21π/2). sin(21π/2)=sin(20π/2+π/2) = sin(10π+π/2) = sin(π/2) = 1.

cos(-37 π/6) = cos(37π/6) = cos(36π/6+π/6) =cos(6π+π/6) = cos(π/6) = . 3 /2

PYTHAGOREAN IDENTITIES. sin2t + cos2t = 1, tan2t +1 = sec2t,cot2t +1 = csc2t.

Proof. We’ve proved the first. To prove the second,multiply both sides by cos2t. For the third, multiply bothsides by sin2t.

Simplify csc t + csc t cot2tsec2t − tan2t

= csc t [1 + cot2t](1 + tan2t) − tan2t

= csc t(csc2t)1

= csc3t

Simplify sin2t − cos2tsin4t − cos4t

= sin2t − cos2t(sin2t − cos2t)(sin2t + cos2t)

= 1

Prove cot θ + tan θ + 1 = cot θ1 − tan θ + tan θ

1 − cot θLet x = tan θ, then cot θ = 1/tan θ = 1/x.Thus we must prove1x + x + 1 = 1/x

1 − x + x1 − 1/x

iff 1x + x + 1 = 1/x1 − x − x2

1 − xiff 1 + x2 + x = [1 − x3]/(1 − x)iff iff (x2 + x + 1)(1 − x) = (1 − x3) 1 − x3 = 1 − x3

iff true

Math 140 Lecture 19Graphs of sin and cosRECALL. sin and cos have period 2π:

sin(x +2 π) = sin(x), cos(x +2π) = cos(x). We often graph periodic functions only over one period, e.g., [0, 2π]. Before and after this interval, they repeat. Graph of sin(x).

π/2 π 3π/2 2π−π/2

1

−1

sin(x)

y=x

x-intercept: x = ..., -2π, -π, 0, ππππ, 2ππππ, 3π, ... .Max value: 1 at x = ..., -3π/2, ππππ/2, 5π/2, ...Min value: -1 at x = ..., -5π/2, -π/2, 3ππππ/2, ...Amplitude: A =1 (see definition below) Period: p = 2π

At x = 0, the line y = x is tangent to the graph of sin(x). Graph of cos(x). Done similarly ... . DEFINITION. The amplitude of a function f is half the

difference between the max and min values of f. Like periods, amplitudes are always positive.

Find the amplitude and period of f, g, and h. f g h

-11

23

2

-2

11

-1

2

3

(3,3)

amplitude = 2 amplitude = 1 amplitude = 1period = 4 period = 4 period = 3

Graph y = 2sin(x) over one period.

amplitude = 2 period = 2π. increases on [0, π/2] and [3π/2, 2π]

For A > 0, y = + Asin(x) has amplitude A. Reflect the graphacross the x-axis for -Asin(x).\

Graph y = sin(2x) over one period. Note sin(x) = 0 iff x = ..., 0, π, 2π, ... . Thus sin(2x) = 0 iff 2x = ..., 0, π, 2π, ... iff x = 0, π/2, π, ... .

Amplitude = 1, period = π. Increases on [0, π/4] and [3π/4, π].For B > 0, y = sin(Bx) has period is 2π/B. Reason: sin(Bx)

repeats at Bx = 2π, thus at x = 2π/B.

THEOREM. For y = + Asin(Bx) & y = + Acos(Bx) with A,B >0, amplitude : A period : p = 2π/B

Graph y = -3cos(πx) over one period. List the amplitude,period, x-intercepts and the intervals in the period onwhich the function increases. y = -3cos(πx) = − Acos(Bx)A = 3, B = π amplitude: A = 3 period: p = 2π/B = 2π/π = 2 x-intercepts? cos(x) = 0 when x = ..., π/2, 3π2, ... .-3cos(πx) = 0 when πx = ..., π/2, 3π/2, ... . iff x = ..., 1/2, 3/2, ... .

3

1

2

-1

-2-3

1/2 1 3/2 2

increases on: [0,1].

Find an equation for the graph. Write it in the form y = + Asin(Bx) or y = + Acos(Bx) with A, B > 0.

(2,2)

The graph has the shape of y = -Acos(Bx). amplitude: A = 2; period: p = 4.B: p = 2π/B ˆ 4 = 2π/B ˆ 4B = 2π ˆ B = π/2.

Equation: y = −2 cos(π2 x)

π/2 π 3π/2 2π

1

-1

sin(x)

2sin(x)

-2

2

π/2 π 3π/2 2π

sin(x)-1

1

sin(2x)

Math 140 Lecture 20 RECALL. For C > 0, the graph of f (x + C) is the graph of f (x) shifted C units to the left. f (x− C) is the graph of f (x) shifted C units to the right. The phase shift is the amount C subtracted from x.f (x+C) = f (x−(-C)) has phase shift -C.f (2x−C) = f ( 2(x−C/2) ) has phase shift C/2.Graph y = sin(x − π/2). Phase shift = π/2.

sin(x) sin(x- /2)π

1

−1

1

−1

π 2π

π/2π/2−π/2

π 2π 5π/2

To graph y = Asin(Bx+D): Factor out B, and rewrite in the form Asin[B(x−C)]:

Asin[Bx+D)] ƒ Asin[B(x+(D/B))] ƒ Asin[B(x−C)]. Graph y = sin(Bx), stretch vertically to get y = Asin(Bx).Shift horizontally to get y = Asin[B(x−C)].

Graph y = sin(2x − π) over one period. sin(2x− π) = sin[2(x− π/2)]. Phase shift = π/2

π is not the shift for sin(2x-π). In sin(2x-π), π is subtracted from 2x.But the shift must be subtracted from x. You must factor sin(2x-π)to sin(2(x-π/2)) to see that the shift subtracted from x is π/2.

Graph -2cos(2x− π/2) over one period. -2cos(2x− π/2) = -2cos[2(x− π/4)] τ amplitude = 2, period = π, phase shift = π/4.

Graph y = csc(x). csc(x) = 1/sin(x). To graph csc(x), graph sin(x) and then invert its values.sin(x) = ½ ˆ csc(x) = 1/(½) = 2. sin(x) = 0 ˆ csc(x) = 1/0 = undef.

x-intercepts: none vert. asymptotes: ..., x = 0, x = π, x = 2π, ... Graph y = tan(x).

tan(x) = sin(x)/cos(x). tan(x) = 0 iff sin(x) = 0 iff x = ..., 0, π, 2π, ... .

x-intercepts: ..., 0, π, 2π, ... . tan(x) undef. iff cos(x) = 0 iff x = ..., -π/2, π/2, 3π/2, ...

vert. asymptotes: ..., x = -π/2, x = π/2, x = 3π/2, ... .

π/2

π 2π0

−π/2

Graph y = -tan(3x + π/2). -tan(3x + π/2) = -tan[3(x− (- π/6))]. Phase shift = -π/6.

Graph tan(x), tan(3x), -tan(3x), then shift left by π/6.

x-intercepts: ..., -π/6, π/6, ... Period = π/3. vert. asymptotes: ..., x = -π/3, x = 0, x = π/3, x = Ν2π/3, ... Note: tan(x) has period π, thus tan(Bx) has period π/B.Graph y = -csc(3x + π) over one period.

sin(2x) sin(2(x- /2))π

ππ/2

π/2

π 3π/2

1

−1

1

cos(x) cos(2x) -2cos(2(x- /4)) π

1

−1

π 2π

-2cos(2x)

2

−2

π

5π/4

π/4 π/4

π/2 π/2

0 π/2

π

2π

csc(x)

1

−1

000

-tan(3x)tan(3x)tan(x)

π/6π /6π /2−π/2 −π/6−π/6

0

-tan[3(x+ /6)]

π/3

2π/3

−π/3

π/6−π/6

π

Math 140 Lecture 21RECALL. sin2+cos2 = 1, sin(-x) = -sin(x), cos(-x) = cos(x).

cos(π/2−x) = sin(x), sin(π/2−x) = cos(x).TRIGONOMETRIC ADDITION FORMULAS.

sin(s ± t) = sin s cos t ± cos s sin tcos(s + t) = cos s cos t − sin s sin tcos(s − t) = cos s cos t + sin s sin ttan(s + t) = tan s + tan t

1 − tan s tan t

tan(s − t) = tan s − tan t1 + tan s tan t

Proof of .cos(s + t) = cos s cos t − sin s sin t

st

s+t

t

-s

(cos(s+t), sin(s+t))

(1,0)

(cos(t), sin(t))

(cos(-s), sin(-s))

dd

1

1

1

1

=(cos(s), -sin(s))

In both pictures the obtuse angle is s +t, the short sides areradii of length 1, and the long sides have some commonlength d. Use the distance formula to calculate d.

In the first,d = (cos(s + t) − 1)2 + (sin(s + t) − 0)2

τd2 = cos2(s + t) − 2 cos(s + t) + 1 + sin2(s + t)τ ,d2 = 2 − 2 cos(s + t)

since .cos2(s + t) + sin2(s + t) = 1In the second,

d = (cos t − cos s)2 + (sin t + sin s)2

τd2 = cos2t − 2 cos t cos s + cos2s+sin2t + 2 sin t sin s + sin2s

τ ,d2 = 2 − 2 cos t cos s + 2 sin t sin ssince cos2t + sin2t = 1, cos2s + sin2s = 1,cos(−s) = cos(s), sin(−s) = −sin(s).

Equate the two terms which equal d 2:

2 − 2 cos(s + t) = 2 − 2 cos t cos s + 2 sin t sin sτ−2 cos(s + t) = −2 cos t cos s + 2 sin t sin sτ .cos(s + t) = cos s cos t − sin s sin t

Proof of .cos(s − t) = cos t cos s + sin t sin scos(s − t) = cos(s + (−t))

= cos s cos(−t) − sin s sin(−t).= cos s cos t + sin s sin t

Proof of .sin(s + t) = sin s cos t + cos s sin tsin(s + t) = cos(π

2− (s + t)) = cos((π

2− s) − t)

= cos(π2

− s)cos t + sin(π2

− s)sin t.= sin s cos t + cos s sin t

The other proofs are similar.

In each problem below and in homework: (1) Apply an addition formula. first point(2) Simplify. second point

cos π5

cos π30

+ sin π5

sin π30

= first pointcos(π5

− π30

)

= second pointcos(6π30

− π30

) = cos(π6

) = 3 /2

sin(x + y)cos y − cos(x + y)sin y

= = sin((x + y) − y) sin x

sin(a + b) + sin(a − b)

= (sin a cos b + cos a sin b) + (sin a cos b − cos a sin b)

= 2 sin a cos b

cos(a − b) − cos(a + b)

= (cos a cos b + sin a sin b) − (cos a cos b − sin a sin b)

= 2 sin a sin b

tan π3 − tan π

6

1 + tan π3 tan π

6

= tan(π3

− π6

) = tan(π6

) = sin(π/6)cos(π/6)

= 1/23 /2

= 13

2 tan(π

8 )1 − tan2(π

8 )

= tan(π8 + π

8 ) = tan(π4 ) = 1

Given . tan s = 13, tan t = 2

3, find tan(s + t)

tan(s + t) = tan s + tan t1 − tan s tan t

= = 13 + 2

3

1 − 13 ⋅ 2

3

= 11 − 2

9

97

Use an addition formula to find tan(π/12). π/12 = 4π/12 − 3π/12 = π/3 − π/4

tan( π12) = tan(π

3 − π4 )

= = = tan(π/3)−tan(π/4)1+tan(π/3)tan(π/4)

3 −1

1+( 3 )(1)

( 3 −1)

(1+ 3 )⋅ (1− 3 )

(1− 3 )

= 2 3 −1−31−3 = 2 3 −4

−2 = 2 − 3

Math 140 Lecture 22DOUBLE-ANGLE FORMULAS.

sin 2θ = 2 sin θcos θcos 2θ = cos2θ − sin2θtan 2θ = 2 tan θ

1 − tan2θProof.

sin 2θ = sin(θ + θ) ,= sin θcos θ +cos θsin θ = 2 sin θcos θ

cos 2θ = cos(θ + θ) ,= cos θcos θ − sin θsin θ = cos2θ − sin2θ

tan 2θ = tan(θ + θ) ,= tan θ + tan θ

1 − tan θ tan θ = 2 tan θ1 + tan2θ

HALF-ANGLE FORMULAS.

+ if θ/2 ι quadrant I or II, − if not.sin θ2 = ± 1 − cos θ

2

+ if θ/2 ι quadrant I or IV, − if not.cos θ2 = ± 1 + cos θ

2tan θ

2= sin θ

1 + cos θProof.

. Hencecos2θ + sin2θ = 1 andcos2θ = 1 − sin2θ

sin2θ = 1 − cos2θ

cos 2θ = cos2θ − sin2θ= (1 − sin2θ) − sin2θ = 1 − 2 sin2θ

τ 2 sin2θ = 1 − cos 2θ

τ . Replacing θ by θ/2 gives sin θ = ± 1 − cos 2θ2

,sin θ2 = ± 1 − cos θ

2

cos 2θ = cos2θ − sin2θ = cos2θ − (1 − cos2θ) = 2 cos2θ −1

τ 2 cos2θ = 1 + cos 2θ

τ . Replacing θ by θ/2 gives cos θ = ± 1 + cos 2θ2

,cos θ2

= ± 1 + cos θ2

tan θ2

= sin θ2

/ cos θ2

= 1 − cos θ2

/ 1 + cos θ2

= 1 − cos θ1 + cos θ

= (1 − cos θ)(1 + cos θ)(1 + cos θ)(1 + cos θ)

,= 1 − cos2θ(1 + cos θ)2 = sin2θ

(1 + cos θ)2 = sin θ1 + cos θ

, . cos θ = 13

3π2 < θ < 2π

Evaluate sin(2θ), cos(2θ), sin(θ/2), cos(θ/2). First find sin(θ). The inequality ˆquadrant IVˆsin θ<0.

sin θ = ± 1 − cos2θ = − 1 − 19 = − 8

9 = −23 2

sin 2θ = 2 sin θcos θ = 2 −2

3 2

13 = −4 2

9

cos 2θ = cos2θ − sin2θ = 19 − 8

9 = −79

ˆ 3π2 < θ < 2π 3π

4 < θ2 < π

ˆ is II quadrant ˆ θ2 sin θ

2 > 0, cos θ2 < 0

sin θ2 = ± 1−cos θ

2 = 1−1/32 = 3−1

6 = 1/ 3

cos θ2 = ± 1+cos θ

2 = − 1+1/32 = − 3+1

6 = − 23

Evaluate using half-angle formulas. = = sin π

12 sin π/62

1 − cos π6

2 = 1 − 3 /22 = 2 − 3

4 = 12 2 − 3

= = cos π12 cos π/6

2

1 + cos π6

2= 1 + 3 /2

2= 2 + 3

4= 1

22 + 3

= = tan π12 tan π/6

2

=sin π

6

1 + cos π6

= 1/21 + 3 /2

= 12 + 3

= 2 − 3(2 + 3 )(2 − 3 )

= 2 − 3

For the angle t in the picture, find cos(2t) and sin(t/2).

t2

3x

First find the missing side x, then find sin(t ) and cos(t).x2 + 4 = 9, x = 9 − 4 = 5

sin t = 5 /3, cos t = 2/3 cos 2t = cos2t − sin2t = 4

9 − 59 = −1

9

sin t2 = ± 1−cos t

2 = 1−2/32 = 3−2

6 = 16

Express in a form which does not use powers ofsin4θtrigonometric functions.

sin4θ = (sin2θ)2 =

1 − cos 2θ2

2= 1 − 2 cos 2θ +cos22θ

4

=1 − 2 cos 2θ + 1

2(1 + cos 4θ)4

= 18(3 − 4 cos 2θ +cos 4θ)

The calculation for is similar. cos4θ

Math 140 Lecture 23 Exam 4 next weekProduct-to-sum rules RECALL

cos(x − y) = cos x cos y + sin x sin y cos(x + y) = cos x cos y − sin x sin y

Adding these gives cos(x − y) + cos(x + y) = 2 cos x cos y

Subtracting gives cos(x − y) − cos(x + y) = 2 sin x sin y

Similarly sin(x − y) = sin x cos y − cos x sin y sin(x + y) = sin x cos y + cos x sin y

Adding these gives sin(x − y) + sin(x + y) = 2 sin x cos y

Dividing by 2 gives PRODUCT-TO-SUM FORMULA.

sin x sin y = 12 [cos(x − y) − cos(x + y)]

cos x cos y = 12 [cos(x − y) + cos(x + y)]

sin x cos y = 12 [sin(x − y) + sin(x + y)]

Write as a sum or difference of trigonometric functions. sin π

2 cos π6

= 12 [sin(π

2 − π6 ) + sin(π

2 + π6 )]

.= 12(sin π

3 + sin 2π3 )

cos x cos 2x = 1

2 [cos(x − 2x) + cos(x + 2x)]. = 1

2 [cos(x) + cos(3x)]

Solving trigonometric equations RECALL. . sin x = sin(π −x), cos x = cos(−x)Note: are called supplementary angles.x and π −xFind all solutions for each equation. sin θ = 1

2

The two simplest solutions: θ = π/6, and θ = π-π/6 = 5π/6.Since sin(θ) has period 2π, adding a multiple of 2π to

either of these also produces a solution. The general solution consists of the two sets θ = π/6 + 2πn or θ = 5π/6 + 2πn for n an integer.

Here we use x for the angle instead of θ.cos x = 12

The two simplest solutions are x = π/3 and x = -π/3. The general solution is

x = π/3 + 2πn or x = −π/3 + 2πn for n an integer. CONVENTION. Assume n is an arbitrary, possibly negative,

integer. We’ll omit the phrase “for n an integer”.

tan x = 13

Since tan π6 = sin(π

6 )/ cos(π6 ) = (1

2)/( 32 ) = 1

3

π/6 is one solution. Since tan(x) has period π, adding amultiple of π also gives a solution. The general solutionis x = π/6 + πn .

For sin, cos, add 2πn to the (usually) two simplestsolutions. For tan, add πn to the one simplest solution.

sin 2x = 1iff 2x = π/2 + 2πn, Only one simple solution here.iff x = π/4 + πn .

RECALL. sin(x) and cos(x) are always between -1 and 1. cos x = 3

iff never. τno solution. 2 cos2x + cos x = 1

2 cos2x + cos x − 1 = 0 (2 cos x − 1)(cos x + 1) = 0

cos x = 12 or cos x = −1

x = π/3 + 2πn or x = -π/3 + 2πn or x = π + 2πn . cos2x − cos x = 2

cos2x − cos x − 2 = 0 (cos x + 1)(cos x − 2)

The second is impossible.cos x = −1 or cos x = 2x = π + 2πn .

2 tan2x − 3 tan x sec x − 2 sec2x = 0 2 sin2x − 3 sin x − 2 = 0

(2 sin x + 1)(sin x − 2) = 0The second is impossible.sin x = −1

2 or sin x = 2x = -π/6 + 2πn or x = -5π/6 + 2πn .

RECALL. A function is 1-1 iff no horizontal line crosses itsgraph more than once.

sin, cos and tan are not 1-1. But they are 1-1 on the heavily marked intervals. These are the largest such intervals containing first quadrant angles.

−π/2 π/2

0

π

sin

cos

tan−π/2 π/2

For sin, the interval is [-π/2, π/2]. For cos, the interval is [0, π]. For tan, the interval is (-π/2, π/2).

y=1/2π/6π−π/6=5π/6

Math 140 Lecture 24Exam 4: Lectures 19-24, including this lecture.

Inverse trigonometric functionsRECALL. A function is 1-1 iff no horizontal line crosses its

graph more than once. While not 1-1 in general, sin, cos, and tan are 1-1 on the

first quadrant angles in [0,π/2] and on the largerheavily marked “1-1” intervals.

−π/2 π/2

0π

sin

cos

tan−π/2 π/2

1

−1

DEFINITION. sin -1, cos

-1, tan -1 are the inverses of the above

restrictions of sin, cos, and tan. Thus sin

-1(x) = the θι[-π/2, π/2] such that sin(θ) = x, cos

-1(x) = the θι[0, π] such that cos(θ) = x, tan

-1(x) = the θι(-π/2, π/2) such that tan(θ) = x. sin-1(x) and cos-1(x) have domain [-1, 1], for x2[-1, 1], they

are undefined. tan -1(x) has domain (-5,5).

NOTATION. sin -1(x), cos

-1(x) and tan -1(x) are also written:

arcsin(x), arccos(x), and arctan(x). Warning, sin

-1(x) = 1/sin(x). sin -1(x) is the inverse;

(sin(x)) -1 = 1/sin(x) is the reciprocal.

sin -1(0) = 0 cos -1(1) = 0 tan -1(0) = 0sin -1(1/2) = π/6 cos -1(λ3/2) = π/6 tan -1(1/λ3) = π/6

sin -1(1/λ2) = π/4 cos -1(1/λ2) = π/4 tan -1(1) = π/4sin -1(λ3/2) = π/3 cos -1(1/2) = π/3 tan -1(λ3) = π/3

sin -1(1) = π/2 cos -1(0) = π/2

The unit circle pictures are

The heavily marked half circles are the restricted ranges.

arcsin(-1/2) = -π/6 arccos(-1/λ2) = 3π/4 cos -1(2) = undef since 22[-1,1] arctan(-1) = -π/4 cos(tan-1(-1)) = cos(-π/4) = cos(π/4) = 1/ 2

Since inverses undo each other, we have sin-1(sin θ) = θ if θι[-π/2, π/2] sin(sin-1

x) = x if xι[-1, 1]cos

-1(cos θ) = θ if θι[0, π] cos(cos-1 x) = x if xι[-1,1]

tan-1(tan θ) = θ if θι(-π/2,π/2) tan(tan -1

x) = x for any x

RECALL. . cos(−x) = cos x, sin(π −x) = sin xWhen a formula involves an angle outside the restricted

range, rewrite it in terms of an angle θ in the restrictedrange before applying the equations in the box.

cos(cos−1(1/5)) = 1/5 cos−1(cos(π/5)) = π/5

�important cos−1(cos(−π5 )) = cos−1(cos(π

5 )) = π5

�important sin−1(sin 4π5 ) = sin−1(sin π

5 ) = π5

THEOREM. For xι[-1,1], cos(sin−1x) = 1 − x2

sin(cos−1x) = 1 − x2

Proof of 1st. Recall: . cos2θ = 1 − sin2θτ cos θ = ± 1 − sin2θτ cos(sin−1x) = ± 1 − (sin(sin−1x))2

= 1 − x2

The “+” was chosen over the “−” since sin-1(x)ι[-π/2, π/2] ˆ cos(sin-1(x))>0.

sin(sin−1(15)) = 1

5

cos(sin−1(15)) = 1 − 1

25 = 2425 = 2 6

5

tan(sin−1(15)) =

sin(sin−1 15)

cos(sin−1 15)

= 1/52 6 /5

= 612

sec[sin−1(−1) − cos−1(1)] = sec[ −π

2 − 0] = 1/ cos(−π2 ) = 1/0 = undef.

In a triangle with hypotenuse 1 and side x, = angle opposite x, sin−1x = angle adjacent to x. cos−1x

x

arcsin(x)

arccos(x)

1

Note that sin−1x + cos−1x = π/2 (= 90o)

xarcsin(x)

arccos(x)

x

x1

-1

-1 1

π/2

−π/2

π 0

Math 140 Lecture 25Trigonometric word problems(a) Find the exact answer. (b) Find the decimal answer.To test if your calculator is in radian or degree mode, calculate sin(1).

sin(1o) ≈ .017, sin(1rad) ≈ .84 . Give only exact answers on tests. The bookstore has cheap trig calculators for less than $20.There are tables for sin and cos on pages A-34 to A-40.Point P is level with the base of a 1000 ft building which is 4000

ft away. Find the angle of elevation from P to the top of thebuilding. (a) the exact radian answer, (b) to nearest degree.

1000

P 4000

θ

Let θ be the angle. tan(θ) = 1000/4000 = 1/4. (a) θ = tan-1(1/4) � exact answer. (b) θ = 14o

� to nearest degree. An antenna sits atop a 1000 ft building. From point a P on the

ground, the angle of elevation to the top of the building is ββββ,,,,

the angle of elevation to the top of the antenna is αααα. Expressthe height of the antenna in terms of αααα and ββββ.

Let h be the height of the antenna.Let x be the distance between P and the building. Thentan β = 1000

x x = 1000

tan β

tan α = h+1000x

x tan α = h + 1000h = x tan α − 1000h = 1000

tan β tan α − 1000Answer: Height is ft. Remember the units1000( tan α

tan β − 1)

From a point P level with the base of a mountain, the angle ofelevation of the mountain is αααα. . . . From a point Q 1 mile closer tothe mountain’s base, the angle of elevation is ββββ. Express theheight of the mountain in terms of α α α α and ββββ.

α β

1

h

xP QLet h be the height. Let x be as pictured.

τ so tan β = hx x = h

tan β

, so tan α = h1+x = h

1+h/ tan βh

1+h/ tan β = tan α

h = (1 + htan β)tan α

h tan β = (tan β +h)tan αh tan β = tan β tan α + h tan αh tan β − h tan α = tan β tan αh(tan β − tan α) = tan β tan α

Answer height = milestan β tan α/(tan β − tan α)

ΛTHEOREM. The area of a triangle with sides a and b andincluded angle θ is ½absin(θ).

PROOF. Case θ is acute.

To find the area, we need to know the height h. sin θ = h

ah = a sin θ

Area = 12hb = 1

2(a sin θ)b = 12ab sin θ

Case θ is obtuse.Recall sin(π−θ) = sin θ.

sin(π − θ) =ha

h = a sin(π − θ) =a sin θ

Area (same answer)= 12hb = 1

2(a sin θ)b = 12ab sin θ

Find the area of an octagon (stop sign) of radius 1 ft.Give the exact answer and the decimal answer to 2 places.

1θ

1

In each of the 8 triangles, the two sides are radii andhave length 1. θ = 2π

8 = π4

The area each triangle = 12(1)(1)sin(π

4 ) = 12( 2

2 ) = 24

The area of the octagon = 8ξthe area of each triangle = 8 2

4 = 2 2Exact answer = sq.ft. Decimal answer = 2.83 sq. ft.2 2

1000

P xβ

α

h

a

bθ

h

a

b

θhπ−θ

Math 140 Lecture 26CONVENTION. Assume side a is opposite angle A, side b is

opposite angle B and side c is opposite angle C. SINE LAWS. In any triangle, the ratio of one angle’s sine and

its opposite side equals the ratio of any other angle’ssine and opposite side.

Although written as one, there are 3 equations. Each involves twosides and two angles.

orsin Aa = sin B

b = sin Cc

asin A

= bsin B

= csin C

PROOF. Recall that the area of a triangle is half the productof any two sides times the sine of the included angle.

Thus the area of the triangle can be written three ways: 1

2bc sin A = 12ac sin B = 1

2ab sin Cmultiply by 2

bc sin A = ac sin B = ab sin Cdivide by abc

sin Aa = sin B

b = sin Cc

COSINE LAWS. For any two sides of a triangle, the sum oftheir squares minus twice their product times the cos ofthe included angle equals the square of the third side.

Each involves three sides and one angle. a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2ac cos B

c2 = a2 + b2 − 2ab cos C

PROOF. We prove the last equality for the case C acute.

C

B

A

c

b

a

x

h

b-x

sin C = ha , cos C = x

a , so h = a sin C, x = a cos C c2 = (b − x)2 + h2

c2 = b2 − 2bx + x2 + h2

c2 = b2 − 2b(a cos C) + a2cos2C + a2sin2C c2 = b2 − 2ab cos C + a2(cos2C + sin2C)

c2 = a2 + b2 − 2ab cos C

STRAIGHT ANGLE SUM FACT. The sum of a triangle’s 3 anglesis a straight angle.

ςA +ςB +ςC = 180o = πIf you know two angles, you can solve for the third.

Solving for C gives ςC = 180o − (ςA +ςB).

Today’s problems involve 4 quantities; each is a side oran angle whose measure is given or wanted. Usually --

For 2 sides and 2 angles: use the sine law involving the2 sides (if necessary, get the third angle with theStraight Angle Sum Theorem). For 3 sides and 1 angle: use the cosine law involvingthe angle.

Given b,

ςA, ςC: find c. 2 side, 2 angle problem. Use the sine law with b and c.

where B = 180o − (A+C) c

sin C = bsin B

Given a, b, c: find ςC. 3 side, 1 angle problem. Use the cosine law with ςC.

. Solve for cosC then take cos-1.c2 = a2 + b2 − 2ab cos CGive an exact answer and a 2-place decimal answer. ςA = 20o, ςB = 30o, c = 40cm. Find a. 2 angle, 2 side problem with sides a, c. Use the sine law for c and a.

where C = 180o − (ςA +ςB) a

sin A= c

sin CΝΝ = 180o

− (20o + 30o) = 180o

− 50o = 130o. Now solve for a.

asin A = c

sin C

�Exact answer a = c sin Asin C = 40 sin 20o

sin 130o cm

�2-place decimal answer= 17.86 cmAdditional table-user step:

, so sin x = sin(π −x) = sin(180o − x). sin C = sin 130o = sin(180o − 130o) = sin 50o

ςA = 20o, b = 50, c = 60. Find a. 3 side, 1 angle problem with angle ςA. Use the cosine law for ςA.

a2 = b2 + c2 − 2bc cos ASolve for a.

a = b2 + c2 − 2bc cos A = 502 + 602 − 2(60)(50)cos 20o

�Exact answer = 6100 − 6000 cos 20o

�2-place decimal answer= 21.49a = 20, b = 20, c = 30. Find ςA. 3 side, 1 angle problem with angle ςA. Use the cosine law for ςA.

a2 = b2 + c2 − 2bc cos ASolve for . cos A

2bc cos A = b2 + c2 − a2

cos A = (b2 + c2 − a2)/2bc A = cos−1[(b2 + c2 − a2)/2bc]

= cos−1(900/1200) = cos−1(3/4) = 41.41o

b

aC B

Ac

c

ab

C

A

B

b

aC B

Ac

Math 140 Lecture 27A triangle is determined uniquely up to congruence (1) by two sides and an included angle and (2) by two angles and an included side.

However, given two sides and a nonincluded angle, theremay be 0, 1, or 2 triangles.

Suppose ςA = 30o, b =2. Then there is no triangle with a=.5, one triangle with a =1, two triangles with a =1.5,and one triangle with a = 2.5.

A A A A

a=1a=1.5 a=2.5b=2 b=2 b=2 b=2

none one two solutions one solution

a=.5

30 30 30

Suppose two sides and a nonincluded angle are known. When solving for the third side using a cosine law, you

may get an answer of the form . Then there is: s ± tno solution if t <0 or both , s ± t < 0two solutions if t >0 and both , s ± t > 0one solution otherwise.

When solving for a second angle using a sine law, youmay get an answer of the form θ = sin-1t, t > 0. There is: no solution if t >1, two solutions if t <1 and the larger of the two sides isadjacent to the given angle (if one angle is θ, the other π-θ), one solution otherwise.

In a triangle, sin A = ½. What are the possible angles, indegrees, for A? One is A=30o, the other is 180−30=150o.

Is there a triangle in which a = 2, b = 3, and ςA = 60o ? Such a triangle exists iff the third side c exists. Solving for c. a2 = b2 + c2 − 2bc cos A4 = 9 + c2 − 2(3)c cos 60o

0 = c2 − 2(3)12c + 5

c2 − 3c + 5 = 0

c =3± 32−4(1)(5)

2(1) = 3± −112 = undefined

Since c is undefined, the answer is no.Try doing this and the next problem by drawing accurate pictures.

a = 2, b = 2, ς A = 30o. Find ςC if ςB is acute. First find ςB using the sin law, then find ςC.

Thus one answer issin Bb = sin A

a , sin B = ba sin A.

= 30o. TheB = sin−1(ba sin A) = sin−1(2

2 sin 30o) = sin−1 12

other answer is 180o−30o = 150o. The acute angle is 30o.

ςC = 180o−(ςA +ςB) = 180o

−(30o +30o) = 120o

.

Polar coordinatesDEFINITION. The polar coordinates (r, θ) of a point P are

its distance r (radius) from the origin and the angle θbetween the positive x-axis and the line from (0,0) to P.

The usual (x, y) are the rectangular coordinates.

x

yrθ

Note, thisis not theunit circle;it has radius r.

(0,0)

P

Plot the points with polar coordinates (1, π/4), (2,− π/2).From the picture we have

sin θ = yr cos θ = x

r r2 = x2 + y2tan θ = y

xy = r sin θ x = r cos θ r = x2 + y2 θ = tan−1 y

x ** Since tan has period π, θ and arctan(y/x) may differ by a multipleof π. Add π if (x, y) is in quadrant II or III. Here we suppose r > 0.

THEOREM. If a point has rectangular coordinates (x, y) andpolar coordinates (r, θ), then:

and (x, y) = (r cos θ, r sin θ)*. Here r > 0.(r, θ) = ( x2 + y2 , arctan y

x)

NEGATIVE r. is the point on the opposite(−r, θ) (r, θ + π)side of the origin (0,0) as . (r, θ)

(r, )

(-r, )θ

θ

(0,0)θ

Convert from polar coordinates to rectangular: .(7, π6 )

r = 7, θ = π/6, (x, y) = (rcosθ, rsinθ) =

�answer (7 cos π6 , 7 sin π

6 ) = (7 32 , 7

2)

Convert from rectangular coordinates to polar: (-1, ).3x = -1, y = . *Point is in quadrant II, τadd π. 3(r, θ) = ( x2 + y2 , tan−1 y

x)

ƒ∗ �answer = ( 1 + 3 , tan−1 3−1 ) = (2, −π

3 ) (2, 2π3 )

Convert the polar equation to a rectangular equation:. r sin θ +2 cos θ = 0

ƒ y + 2(x/r) = 0 y + 2 xx2+y2

= 0

� answer (hw simplifies more)x2 + y2 y + 2x = 0

Convert the rectangular equation to a polar equation: 2x − y2 = 0

� answer2r cos θ − r2sin2θ = 0

side-angle-side angle-side-angle two sides, nonincluded

Math 140 Lecture 28The graph of y = ax2+bx+c is a vertical parabola with a

vertical axis of symmetry. Other parabolas havehorizontal or slanted axes.

DEFINITION. A parabola consists of all points equidistantbetween a given focus point and a given directrix line.The axis is the line through the focus and perpendicularto the directrix. The vertex is the intersection of theparabola and the axis.

In a parabola light rays parallel to the axis are reflected tothe focus. Telescope mirrors and satellite antennas havethis shape. The vertex lies halfway between the focusand the directrix.

Find the equation for the parabola with focus (0, p) anddirectrix y = -p.

(0,p) = focus

y=-p

(x,y)

directrix

(0,0) = vertex

axis

y

p

For any point (x, y), The distance between (x, y) and the directrix y = -p is y+p.The distance between (x, y) and the focus (0, p) is

. (x − 0)2 + (y − p)2 = x2 + y2 − 2py + p2

(x, y) is on the parabola iff the distances are equal iff y + p = x2 + y2 − 2py + p2

iff (y + p)2 = x2 + y2 − 2py + p2

iff y2 + 2py + p2 = x2 + y2 − 2py + p2

iff iff 2py = x2 − 2py 4py = x2

iff iff x2 = 4py x2 = ky where k = 4p and p = k/4

VERTICAL PARABOLA THEOREM. For k = 0, the graph of is the vertical parabola with focus (0, p) andx2 = ky

directrix y = -p where p = k/4. The axis is the y-axis; thevertex is (0, 0).

Exchanging x and y gives ∞ HORIZONTAL PARABOLA THEOREM. For k = 0, the graph of

is a horizontal parabola with focus (p, 0) and y2 = kxdirectrix x = -p where p = k/4. The axis is the x-axis; thevertex is (0, 0).

x2-parabolas such as y = x2 are vertical; y

2-parabolas are horizontal. Find the focus, directrix and graph of y = -x2/8.

x2 = -8y. Parabola is vertical

p = k/4 = -8/4 = -2. Focus (0, p): = (0, -2). Directrix y = -p: y = 2. Graph ... .

Hint, first mark the two focal-width points on a line through the focus,|| to the directrix and equidistant from focus and directrix.

Find the focus, directrix and graph of . 3y2 = 4xWrite it in horizontal-parabola form: y

2 = kx.

. y2 = 43x, p = k/4 = 4

3 /4 = 13

Focus (p, 0): (ƒ, 0). Directrix x = -p: x = -ƒ. Graph ... .

THEOREM. In any equation, replacing each x by x−a shifts the graph right by a unitsx by x+a shifts the graph left by a unitsy by y−b shifts the graph up b unitsy by y+b shifts the graph down b units

When a parabola is shifted, so are its focus, directrix,vertex, and axis.

To graph a parabola, get the squared variable on the left,the rest on the right. Complete the square if needed.Write the equations in the form:

or (x ± a)2 = k(y ± b) (y ± b)2 = k(x ± a)

Find the focus, directrix and graph of .x2 − 2x + 9 − 8y = 0, �complete the squarex2 − 2x = 8y − 9, x2 − 2x + 1 = 8y − 8

, . A vertical parabola.(x − 1)2 = 8(y − 1) p = k/4 = 8/4 = 2For : vertex = (0,0), focus = (0,2), directrix y = -2. x2 = 8yTo get , shift right 1 unit and up 1 unit.(x − 1)2 = 8(y − 1)Shifting (0,0) right 1 and up 1 gives the vertex: (1,1). Shifting (0,2) gives the focus: (1,3). Shifting y = -2 gives the directrix: y = -1.

directrix: y=-1

focus= (1,3)

vertex= (1,1)

axis

focus

directrix

parabola

vertex

(p,0) = focus

x=-pdirectrix

4px=y2

axis(0,0) vertex

focal-width point

Math 140 Lecture 29Find the focus, directrix and graph of .y2 + 2y = 4x − 5

y2 + 2y + 1 = 4x − 4, (y + 1)2 = 4(x − 1). 4p = 4, p = 1

This is the parabola y2 = 4xshifted: down 1 unit, right 1 unit.The y

2 means the parabola is horizontal.Shifting (0,0) down 1 and right 1 gives the vertex: (1,-1).Shifting (p,0) = (1,0) gives the focus: (2,-1).Shift x= -p, i.e., x=-1, gives the directrix: x=0.

directrix: x=0

focus=(2,-1,)vertex= (1,-1)

RECALL. A circle is the set of all points such that thedistance to a center point is some constant r.

focifoci

minor axis

major axis

vertices

center

DEFINITION. An ellipse is the set of all points such that thesum of the distances to two focus points is someconstant. The major axis goes from a vertex at one endof the ellipse through the two foci to the vertex at theopposite end. The minor axis is a perpendicular bisectorof the major axis.

The vertices are the two points farthest apart. The majorradius a is half the major axis length; the minor radiusb is half the minor axis length; the focal radius is halfthe distance between the foci.

A light ray emitted from one focus point is reflected to theopposite focus point. Planetary orbits are ellipses.

THEOREM. Let a, b, c be the major, minor and focal radii.Then . τ .a2 = b2 + c2 c = a2 − b2

The graph of is a circle with center (0, 0) andx2 + y2 = r2

radius r. This can be written as .x2

r2 + y2

r2 = 1

THEOREM. For a > b > 0, the graphs of andx2

a2 + y2

b2 = 1

are ellipses. (0, 0) is the center. a, b, c arey2

a2 + x2

b2 = 1

the major, minor and focal radii where .c = a2 − b2

is a horizontal ellipse: the foci are (-c, 0) andx2

a2 + y2

b2 = 1(c, 0), the major axis is the line segment (-a,0)(¯¯ a,0), theminor axis is (0,¯ -b)(0¯¯ ,b). Here the bigger # is under x.

is a vertical ellipse: the foci are (0, -c) andy2

a2 + x2

b2 = 1(0, c), the major axis is the line segment (0,-¯¯ a)(0¯¯ ,a), theminor axis is (-b,0)(¯¯ b,0). Here the bigger # is under y.

To graph, complete the squares if necessary. Get 1 on theright. Write the equation in one of the forms above.

Find the major and minor axes, the foci and draw thegraph of .4x2 + y2 = 4

x2 + y2

4 = 1, x2

12 + y2

22 = 1, y2

22 + x2

12 = 1

a = 2, b = 1, c = 4 − 1 = 3Vertical ellipse (y has the bigger denominator).Major axis: (0,-2)¯¯¯¯ (0,2)¯¯¯ , minor axis: (-1,0)¯¯¯¯ (1,0)¯¯¯ , foci: .(0, − 3 ), (0, 3 )

Set your compass to a major radius. Put the point at the end of a minor radius. The twofoci are where the arc of the compassintersects the major axis.

Find the axes and foci. Draw the graph of .16x2 − 96x + 25y2 = 256

complete square for x16(x2 − 6x + ) +25y2 = 25616(x2 − 6x + 9) + 25y2 = 256 + 16 ⋅ 9

divide by 400, get 1 on right16(x − 3)2 + 25y2 = 400(x − 3)2

25 + y2

16 = 1(x − 3)2

52 + y2

42 = 1

a = 5, b = 4, c = 25 − 16 = 9 = 3Horizontal ellipse (x has the bigger denominator).

This is shifted right 3 units.x2

52 + y2

42 = 1Shifting (-5,0)(5,0)¯¯¯¯¯¯¯¯ right 3 gives the major axis: (-2,0)(8,0)¯¯¯¯¯¯¯¯ .Shifting (0,-4)(0,4)¯¯¯¯¯¯¯¯ right 3 gives the minor axis: (3,-4)(3,4)¯¯¯¯¯¯¯¯ .Shifting (-3,0), (3,0) right 3 gives the foci: (0,0), (6,0).

(3,4)

(3,-4)

(-2,0) (8,0)(0,0) (6,0)

focus focus

(1,0)(-1,0)

(0,2)focus

focus(0,-2)

Math 140 Lecture 30

--------

conjugate ax

is

-----------transverse axis

----verte

xvertex---- ---focal axis---focal point

focal point---

-

----asy

mptote

center

horizontal hyperbola

DEFINITION. A hyperbola is the set of all points such thatthe difference of the distances to two focus (or focal)points is some constant. The focal axis through the twofoci intersects the hyperbola at its two vertices. Thetransverse axis is the segment between the vertices, theconjugate axis is a perpendicular bisector. The ends ofthe axes are the midpoints of a box whose diagonals areasymptotes. The box corners and the foci areequidistant from the center of the box. To get the foci,draw an arc around the box center from the box cornerto the focal axis.

The path of a stone thrown upward (in a vertical gravitationalfield) is a parabola. The path of an orbiting planet orcomet is an ellipse with the sun at one focus. The pathof a comet passing through the solar system is one pieceof a hyperbola with the sun at one focal point.

THEOREM. The graphs of and arex2

a2 − y2

b2 = 1 y2

a2 − x2

b2 = 1

hyperbolas (a must be below the positive square). a = the transverse radius = ½ transverse axis length. b = the conjugate radius = ½ the conjugate axis length. c = a2 + b2

= the diagonal radius = ½ the length of a diagonal = the focal radius = ½ the distance between the foci.

is a horizontal hyperbola with x2

a2 − y2

b2 = 1

foci: (-c, 0) and (c, 0)transverse axis: (-a, 0)¯ (a, 0)conjugate axis: (0,¯ -b) (0,¯¯ b)

is a vertical hyperbola with y2

a2 − x2

b2 = 1

foci: (0, -c) and (0, c)transverse axis: (0,¯ -a) (0,¯¯ a)conjugate axis: (-b, 0)¯ (b, 0)

To graph, complete the squares if necessary, then writethe equation in one of the above forms.

Find the axes and foci and draw the graph andasymptotes of . 4y2 − x2 = 4

y2 − x2

4= 1, y2

12 − x2

22 = 1

a = 1, b = 2, c = 1 + 4 = 5Vertical hyperbola (the y

2 is positive).Transverse axis: (0,-1¯¯¯)(0,1¯¯¯ ), conjugate axis: (-2,0¯¯¯)(2,0¯¯¯ ), foci: (0,λ5), (0, -λ5).

focal point

(2,0)

(0,1)

focal point

(0,-1)

(-2,0)

Find the axes and foci and draw the graph andasymptotes of . 16x2 − 96x − 25y2 = 256

complete square for x16(x2 − 6x + ) −25y2 = 25616(x2 − 6x + 9) − 25y2 = 256 + 16 × 9

divide by 400, get 1 on right16(x − 3)2 − 25y2 = 400(x − 3)2

25 − y2

16 = 1

(x − 3)2

52 − y2

42 = 1

a = 5, b = 4, c = 25 + 16 = 41Horizontal hyperbola (the x2 is positive).

This is shifted right 3 units.x2

52 − y2

42 = 1

Shifting (-5,0)(5,0)¯¯¯¯¯¯¯¯ right 3 gives the transverse axis: (-2,0)(8,0)¯¯¯¯¯¯¯¯ .Shifting (0,-4)(0,4)¯¯¯¯¯¯¯¯ right 3 gives the conjugate axis: (3,-4)(3,4)¯¯¯¯¯¯¯¯ .Shifting (0,-λ5), (0,λ5) right 3 gives the foci: (3,-λ5), (3,λ5).

(3,4)

(3,-4)

(-2,0) (8,0)focal point focal point