Lecture 4 - Curve Sketching,Indices and Standard Form

C2 Foundation Mathematics (Standard Track)

Dr Linda Stringer Dr Simon Craikl.stringer@uea.ac.uk s.craik@uea.ac.uk

INTO City/UEA London

Lecture 4 skills

I find the y-intercept, x-intercept(s) and stationary point(s) ofa quadratic or cubic curve

I find the second derivative of an equationI determine the nature of a stationary pointI sketch a quadratic curveI sketch a cubic curve

I apply the rules of indices to simplify expressionsI convert a number to and from standard form

Lecture 4 vocabulary

I interceptI stationary pointI horizontalI maximumI minimumI point of inflectionI nth root, n

√

I standard form

Different types of curves to sketchStraight lines y = mx + c

Quadratic curves y = ax2 + bx + c, where a 6= 0:

Cubic curves y = ax3 + bx2 + cx + d , where a 6= 0:

Sketching curves

When sketching a curve, there are 4 main details you need toshow:

1. The y-intercept. Where the curve crosses the y-axis.

2. The x-intercept(s). Where the curve crosses (or touches)the x-axis.

3. The stationary point(s). Where the curve has gradient 0 (ishorizontal).

4. How the curve behaves for large positive and largenegative values of x.

Stationary points

I Consider a curve with equation y = f (x).I A stationary point of y = f (x) is a point (a, f (a)) where the

curve has gradient 0 (is horizontal).I To find a stationary point you first find dy

dx . You then solve

the equation dydx = 0. The solutions are the x-coordinates of

the stationary points.I To get the y-coordinates, substitute the x-coordinates into

f (x).

Example 1I Question: What is the stationary point of y = x2 − 4x + 7?

2 4

2

4

6

•

I Answer: We have dydx = 2x − 4 (the gradient function).

I Solve dydx = 0.

If 2x − 4 = 0 then x = 2. This is the x-coordinate of thestationary point.

I The y-coordinate is y = 22 − 4× 2 + 7 = 3.I The stationary point of y = x2 − 4x + 7 is (2,3).

Example 2I Question: What are the stationary points of

y = x3 + x2 − 2x?

−2 −1 1

−4−2

24

•

•

I Answer: We have dydx = 3x2 + 2x − 2.

I Solve dydx = 0. It follows 3x2 + 2x − 2 = 0. Using the

quadratic formula we get x = −1±√

73 . These are the

x-coordinates of the two stationary pointsx = 0.55,−1.21 to 2 d.p.

I The y-coordinates arey = (−1±

√7

3 )3 + (−1±√

73 )2 − 2× −1±

√7

3 .y = −0.63,2.11 to 2 d.p.

I The stationary points are (0.55,−0.63) and (−1.21,2.11).

The nature of stationary points

A stationary point is exactly one of the following:

1. Minimum

2. Maximum

3. Point of inflection

I Consider the graph y = f (x).I A stationary point with x-coordinate a is a minimum if for all

points b close to a, we have f (b) > f (a).I A stationary point with x-coordinate a is a maximum if for

all points b close to a, we have f (b) < f (a).I If a stationary point is neither a maximum nor a minimum it

is a point of inflection.

Maxima and minima - example

−2 2 4

−200

−100

100

200

x

y

Point of inflection - example

−2 −1 1 2

−5

5

• x

y

y = x3

Determining the nature of a stationary point

I How do we tell when a stationary point is a minimum,maximum or point of inflection?

I We use the second derivative.

I The derivative dydx is a function of x. This means we may

differentiate it again - to get the second derivative.I For the second derivative we use the notation

d2ydx2 .

I We say ’d squared y by d x squared’

The second derivative - examples

I What is the second derivative of y = x2 − 4x + 7?I Answer: We have dy

dx = 2x − 4.

I Then d2ydx2 = 2.

I What is the second derivative of y = x3 + x2 − 2x?I Answer: We have dy

dx = 3x2 + 2x − 2.

I Then d2ydx2 = 6x + 2.

Determining the nature of stationary points

Consider a stationary point of y = f (x) with x-coordinate a.

I If d2ydx2 (a) > 0 then the stationary point is a minimum.

I If d2ydx2 (a) < 0 then the stationary point is a maximum.

I If d2ydx2 (a) = 0 then we cannot tell.

The nature of the stationary points - example

I What is the nature of the stationary points of the curve withequation y = x3 + x2 − 2x?

I Answer: We have dydx = 3x2 + 2x − 2.

I Then d2ydx2 = 6x + 2.

I We previously found that the stationary points of this curvewere (0.55,−0.63) and (−1.21,2.11) to 2 d.p. We onlyneed to use the x-coordinate to determine whether eachpoint is a maxmum or minimum.

I d2ydx2 (0.55) = 6× 0.55 + 2 = 5.3 > 0 so this is a minimum

I d2ydx2 (−1.21) = 6×−1.21 + 2 = −5 < 0 so this is amaximum

The nature of the stationary points - example

I What is the nature of the stationary point of y = x2 − 1?I Answer: We have dy

dx = 2x

I and d2ydx2 = 2.

I d2ydx2 > 0, so this stationary point is a minimum.

How to sketch a curve y = f (x)

1. Find the y-intercept. Let x = 0. Evaluate y = f (0).

2. Find the x-intercept(s). Let y = 0. Solve f (x) = 0.

3. Find the stationary point(s), and determine the nature ofeach one. Find dy

dx (the derivative or gradient function).

Solve dydx = 0. Find d2y

dx2 (the second derivative). Evaluated2ydx2 for each stationary point.

4. Take a large positive number m and calculate f (m) andf (−m) to see if they are positive or negative.

Curve sketching - example 1

I Sketch the graph y = x2 − 1.I Answer: y = x2 − 1 passes through the y-axis when x = 0.

So y = 02 − 1 = −1.I The graph passes through the x axis when y = 0. Solve

0 = x2 − 1. So x = −1,1.I To work out the stationary points we solve dy

dx = 0. We

have dydx = 2x so x = 0. We need to determine the nature

of the stationary point. Use the second derivative. We haved2ydx2 = 2. Substitute in the x-coordinate of the stationary

point d2ydx2 (0) = 2 > 0 so the stationary point is a minimum.

I When x = 100 the y-value is large and positive. Whenx = −100 the y-value is large and positive.

Sketch of example

Curve sketching - example 2

I Sketch the graph y = x3 + x2 − 2x.I Answer: y = x3 + x2 − 2x passes through the y-axis when

x = 0. So y = 03 + 02 − 2× 0 = 0.I The graph passes through the x axis when y = 0. Solve

0 = x3 + x2 − 2x. So 0 = x(x2 + x − 2) = x(x + 2)(x − 1).It follows x = −2,0,1.

I To work out the stationary points we solve dydx = 0. We

have dydx = 3x2 + 2x − 2 so x = −1±

√7

3 . We need todetermine the nature of the stationary point. Use thesecond derivative. We have d2y

dx2 = 6x + 2. Substitute in the

x-coordinate of the stationary point d2ydx2 (0.55) > 0 so this

stationary point is a minimum. For the otherd2ydx2 (−1.21) < 0 so this stationary point is a maximum.

I When x = 100 the y-value is large and positive. Whenx = −100 the y-value is large and negative.

Sketch of example

Straight lines, y = mx + c

A straight line has gradient m.It has exactly one x-intercept and one y-intercept.

Quadratic curves, y = ax2 + bx + c

A quadratic curve has exactly one stationary point.

A quadratic curve has exactly one y-intercept and zero, one ortwo x-intercepts.

The gradient function is dydx = 2ax + b

Cubic curves, y = ax3 + bx2 + cx + d

A cubic curve has zero, one or two stationary points.

A cubic curve has exactly one y-intercept and one, two or threex-intercepts.

The gradient function is dydx = 3ax2 + 2bx + c

Indices (powers)

xn = x × x × . . .× x

(x multiplied by itself n times)

What is 12, 13, 1100 ?x2 x squaredx3 x cubedxn where n ≥ 4 x to the power n

√x = 2√

x the square root of x3√

x the cube root of xn√

x where n ≥ 4 the nth root of x

Indices

xn = x × x × . . .× x

x−1 = 1/x

x−n = 1/xn = 1/x × x × . . .× x

x0 = 1

What is 2−1 ?What is 2−2, 3−3, 1−50 ?What is 20, 30, 10, 10000, 00 ?

Indices

xn × xm = xn+m

xn ÷ xm = xn/xm = xn × (1/xm) = xn × x−m = xn−m

(xn)m = xnm

Indices

x1/n = n√

x

xn/m = (x1/m)n = ( m√

x)n

xn/m = (xn)1/m = m√

xn

Indices - summary

m,n are real numbers and x can be a number or a variable

x−1 = 1/x

x−n = 1/xn

xn × xm = xn+m

xn ÷ xm = xn/xm = xn × (1/xm) = xn × x−m = xn−m

(xn)m = xnm

x1/n = n√

xxn/m = ( m

√x)n = m

√xn

x0 = 1

Examples

I Simplify x2 × x3

I Answer:x2 × x3 = x2+3

= x5

I Simplify 3x1/2 × 14x3/2

I Answer:

3x1/2 × 14x3/2 = 3× x1/2 × 1

4 × x3/2

= 3× 14 × x1/2 × x3/2

= 34x(1/2+3/2)

= 34x2

Standard form

I Standard form provides a means way to write large orsmall numbers in a short way.

I

A× 10n

Where 1 ≤ A < 10 and n is an integer (n can be 0).

Standard form

I If the number is large count the number of digits to theright of the first digit.

I Express 1500 in standard form:I Answer: 1.5× 103.I Express 2,670,000 in standard form:I Answer: 2.67× 106.I If the number is small count the number of digits to the left of

the first non-zero digit.I Express 0.0035 in standard form:I Answer: 3.5× 10−3.I Express 0.0000007614 in standard form:I Answer: 7.614× 10−7.

More practise with standard form

A× 10n

where 1 ≤ A < 10 and n is an integer (n can be 0)

Express the following in standard form501/2

1/4

5110110

Multiplying in standard form

I Multiplication

(A× 10n)× (B × 10m) = (A× B)× 10n+m

I Question: Express the following in standard form

(3× 104)× (2× 102)

.I Answer:

(3× 104)× (2× 102) = (3× 2)× 104+2

= 6× 106

Multiplying in standard formI Question: Express the following in standard form

(2× 105)× (1× 10−3)

I Answer:

(2× 105)× (1× 10−3) = (2× 1)× 105+(−3)

= 2× 102

I Question: Express the following in standard form

(6× 105)× (4× 108)

I Answer:

(6× 105)× (4× 108) = (6× 4)× 105+8

= 24× 1013

= 2.4× 1014

Dividing in standard formI Division

(A× 10n)÷ (B × 10m) = (A÷ B)× 10n−m

I Question: Express the following in standard form

(4× 106)÷ (2× 103)

I Answer:

(4× 106)÷ (2× 103) = (4÷ 2)× 106−3

= 2× 103

I Question: Express the following in standard form

(2× 105)÷ (8× 108)

I Answer:

(2× 105)÷ (8× 108) = (2÷ 8)× 105−8

= 0.25× 10−3

= 2.5× 10−4