standard 9.0 determine how the graph of a parabola changes as a, b, and c vary in the equation...

Standard 9.0 Determine how the graph of a parabola changes as a, b, and c

vary in the equation

Students demonstrate and explain the effect that changing a coefficient has on the graph of

quadratic functions

GRAPHING A QUADRATIC FUNCTION

A quadratic function has the standard form

y = ax 2 + bx + c where a 0.


The graph is “U-shaped” and is called a parabola.


The highest or lowest point on the parabola is called the ver tex.

What is another word for highest?What is another word for lowest?

In your notes, draw and label the maxima of a parabola.

CORRECT!! The highest point is called maxima And the lowest point is called the minima?


In general, the axis of symmetry for

the parabola is the vertical line

through the vertex.


These are the graphs of y = x 2

and y = x 2.


The y-axis is the axis of symmetryfor both graphs.

Graphing a Quadratic Function

Graph y = 2 x 2 – 8 x + 6

x = – = – = 2 b2 a

– 82(2)

y = 2(2)2 – 8 (2) + 6 = – 2

So, the vertex is (2, – 2).

(2, – 2)

The x-coordinate is:

The y-coordinate is:

Find and plot the vertex.

VERTEX FORM OF A QUADRATIC FUNCTION


CHARACTERISTICS OF GRAPH

Vertex form: y = a (x – h)2 + k

For this form, the graph opens up if a > 0 and opens down if a < 0.

The vertex is (h, k ).

Open up “ + a”Open down “ - a”.

Regular, narrow, wide. “a”

Centered, horizontal shift left, horizontal shift right “h”

Centered, vertical shift up, vertical shift down “k”

Is h positive or negative?

positive move h units to the right

negative move h units to the left

Is a positive or negative?

positive open up

negative open down

Is k positive or negative?

positive move k units up

negative move k units down

Transforming the Graph of a Quadratic Function y=a(x-h)2+k

Quadratic Function8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x2


y = a(x-h)2 + k8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-1f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-3f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-4f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-5f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2-6f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2+0.8

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2+73

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2+4

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x2+5.3

f x = x2


Things to Notice

• Where is the constant term located?

• Outside of the power of 2

• Inside the power of 2

8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-1 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-2 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-3 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-4 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-5 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x-6 2f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+1 2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+2 2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+3.5 2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+5 2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+142

2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = x+283

2

f x = x2


-10 -5 5 10

8

6

4

2

-2

-4

-6

-8

f x = x2


-10 -5 5 10

8

6

4

2

-2

-4

-6

-8

h x = -x2f x = x2


Describe the change from blue to red.

-10 -5 5 10

8

6

4

2

-2

-4

-6

-8

h x = -x2

f x = x2

g x = - x-3 2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = -x2

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = -x2-1

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = -x2-3

f x = x2


Parabola in foci in motion8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

g x = -x2-5

f x = x2


8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f x = x-2 2-3



Graph y = – (x + 3)2 + 412

SOLUTION The function is in vertex form

y = a (x – h)2 + k.

a = – , h = – 3, and k = 4

12

First graph y=x2

4

2

-2

-4

-5 5


Graph y = – (x + 3)2 + 412


y = a (x – h)2 + k.

a = – , h = – 3, and k = 4

12

First graph y=x2

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

Then graph y x2


Graph y = – (x + 3)2 + 412


y = a (x – h)2 + k.

a = – , h = – 3, and k = 4

12

First graph y=x2

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

Then graph y x2

Then graph y (x 3)2


Graph y = – (x + 3)2 + 412


y = a (x – h)2 + k.

a = – , h = – 3, and k = 4

12

First graph y=x2

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

Then graph y x2

Then graph y (x 3)2 Then graph y (x 3)2 4


Graph y = – (x + 3)2 + 412


y = a (x – h)2 + k.

a = – , h = – 3, and k = 4

12

First graph y=x2

4

2

-2

-4

-5 5

Finaly graph y 1

2(x 3)2 4

Then graph y x2

Then graph y (x 3)2 Then graph y (x 3)2 4

4

2

-2

-4

-5 5


Graph y = – (x + 3)2 + 412

The graph of y 1

2(x 3)2 4 is...

4

2

-2

-4

-5 5


Write an equation that could describe the red Parabola

6

4

2

-2

-4

-6

-5 5

y x2



6

4

2

-2

-4

-6

-5 5

y (x 4)2

y x2



y x26

4

2

-2

-4

-6

-5 5



y x2 3

y x26

4

2

-2

-4

-6

-5 5



y x2

10

8

6

4

2

-2

-5 5



y x2

10

8

6

4

2

-2

-5 5

y (x 2)2 4.5



y x22

-2

-4

-6

-8

-10

-5 5



y x2

y x2 6

2

-2

-4

-6

-8

-10

-5 5



y x24

2

-2

-4

-6

-8

-5 5



y x2

y (x 3)2 1.5

4

2

-2

-4

-6

-8

-5 5


What can you conclude about the constantsinside the parenthesis and outside the parenthesis?

2( )h ky x

standard 9.0 determine how the graph of a parabola changes as a, b, and c vary in the equation...

Documents

red parabola

graph y

graph of y

vertex form y

quadratic y

parabola changes

quadratic function y

graphs of y