Our Parabola & Polynomial Goals -- FebruaryMaterial on Quiz and Exam

Student will be able to:

If given Quadratic Function in Standard Form: ID Vertex, Axis of Symmetry, x and y intercepts

Sketch Parabola Rewrite equation into Quadratic Fcn‘s Std Form ID LH and RH Behavior of Polynomials Perform Long Division of Polynomials Perform Synthetic Division of Polynomials

On Tuesday’s Quiz

Standard Form for a Quadratic Function (Parabola)

f(x)=a(x-h)2+k

• hAxis of symmetry, its a vertical line at x=h

• k Vertex at ( h, k ). Note: ‘h’ is (must be) subtracted from

x.• also

a) If a>0 then it opens upwards, if a<0, then it opens downwards.

b) ‘a’ also tells us the “fatness”/ “skinniness” of the parabola

How would we find the x and y intercepts?

Substitute O for x (y intercept) and O for y (x intercept)

e.g. f(x)= 3(x-1)2 - 9• 0= 3(x – 1)2 – 9 9=3(x – 1)2

9/3=(x – 1)2

3=(x – 1)2

±√3 = x – 1

(±√3 +1, 0)x intercept

e.g. f(x)= 3(x-1)2 - 9•f(x)= 3(0-1)2 - 9• f(x)= 3( – 1)2 – 9 f(x)= 3 – 9

f(x)= – 6 ( 0, – 6)y intercept

Can U

Graph it

?

Sketch it…using h, k and intercepts

#s 1 – 8, on page 270

f(x)=5(x-3)2 + 4

f(x)=1(x+3)2 – 4

Then graph these

Let’s do problems in book…page 270, #s 13, 14, 16, 17

If we’re given “h” and “k”, and a point on the parabola…We can write its function, f(x)

e.g., if h=3, and k=5 and the parabola passes thru (0,0)… f(x)=a(x-h) 2+k

• f(x)=a(x-3)2 +5•We can find a by substitution … • 0=a(0-3)2 +5• –5=9a• – 5/9=a f(x)= – 5/9(x-

3)2+5

QUICK REVIEW

The steps to sketching a parabola:1. Put in Standard Form2. Identify the Vertex3. Determine Direction4. Determine where it crosses the y-axis5. Sketch it.

1. f(x)=a(x-h)2+k ____________

2.Vertex ( ___ , ___

)

3.Direction

_________

4.Y intercept ( ____ ,

____ )

WORKSHEET

POP QUIZ

ID the vertex, the direction of the parabola & its y-intercept:

a) f(x) = 5(x – 2)2 + 4

b) f(x) = –5(x – 1)2 + 3

c) f(x) = – 2(x + 3)2 + 4

d) f(x) = (x – 3)2 + 2

Vertex ( ___ , ___ )

Direction ___________

Y intercept ( ____ , ____ )

ANSWER FORMAT

50 Points

POP QUIZ Now Graph each of them.

a) f(x) = 5(x – 2)2 + 4

b) f(x) = –5(x – 1)2 + 3

c) f(x) = – 2(x + 3)2 + 4

d) f(x) = (x – 3)2 + 2

What if NOT in Standard Form?

Put in Std Form by Completing the Square!!When ‘a’ = 1 a(x2+bx)+c•Coefficient in front of x2 must be 1.•take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis above and subtract it as well.•a(x2+bx+ (b/2)2 ) + c − (b/2)2 • This creates a perfect square trinomial (x+b/2)2 + ck oPutting into standard Quadratic Function gives:o (x – − b/2)2 + c , h=− b/2 and k=c

Becomes the c in the standard formula

What if NOT in Standard Form?Put in Std Form by Completing the SquareWhen ‘a’ not equal to 1 ! •Coefficient in front of x2 must be 1, so pull the ‘a’ out

a(x2+bx)+c•take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis and subtract it as well a(x2+bx+ (b/2)2 − (b/2)2 ) + c• Simplify to create a perfect square trinomial

(x+b/2)2 + c − a(b/2)2 oPutting into standard Quadratic Function gives:o (x – − b/2)2 + c , h=− b/2 and k=c

Becomes the c in the standard formula

What if NOT in Standard Form?Put in Std Form by Completing the SquareWhen ‘a’ not equal to 1 ! •Coefficient in front of x2 must be 1, so pull the ‘a’ out

a(x2+bx)+c•take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis and subtract it as well a(x2+bx+ (b/2)2 − (b/2)2 ) + c• Simplify to create a perfect square trinomial

(x+b/2)2 + c − a(b/2)2 oPutting into standard Quadratic Function gives:o (x – − b/2)2 + c , h=− b/2 and k=c

Becomes the c in the standard formula

Problems page 271,#s 29-34

Real Life Applications-Gallery Walk

page 272, #s 78-84

•Groups of four with these roles:o Solver – math problem solution leadero Recorder – easel pad leadero Speaker during Gallery Walko Time Keeper• Time tables are on the white board !• So is the format of the expected easel chart !

Simple Example… from class

1x2 – 2x

1 [ 1x2–2x +(2/2)2 – (2/2)2] + 0

1[x2–2x +(1)2 ] – 1 (1)2 + 0

[x–1]2 – 1

Real Life Applications - Gallery Walk

Example… Problem #78

–.008x2+1.8x + 1.5

–.008 [ 1x2–225x +(112.5)2 – (112.5)2]

+ 1.5

–.008[x2–225x +(112.5)2 ] + (–.008) (–

(112.5)2 + 1.5

–.008[x–112.5]2 + (–.008)(– (112.5)2 +

1.5

–.008[x–112.5]2 + 102.75

What if it is written in the standard form of a quadratic equation…

OR Memorize

for ax2+bx+c, memorize:• Vertex is [ –b/2a, f(–b/2a) ]• Axis of symmetry is at x= –b/2a