# graphing quadratic functions

DESCRIPTION

Graphing Quadratic Functions. Graphs of Quadratic Functions. Axis of symmetry. Important features of graphs of parabolas. x-intercepts. Vertex. Graphing Quadratics. If you were asked to graph a quadratic, what information would you need to know to complete the problem?. - PowerPoint PPT PresentationTRANSCRIPT

Graphing Quadratic Functions

Graphs of Quadratic Functions

Vertex

Axis of symmetry

12

10

8

6

4

2

-2

-10 -5 5 10

g x = -0.5x+4 2+6

f x = 2x-2 2+1

x-intercepts

Important features of graphs of parabolas

Graphing QuadraticsIf you were asked to graph a quadratic, what information would you need to know to complete the problem?

The vertex, because we need to know where the graph is located in the plane

If the parabola points up or down, and whether it opens normal, narrow or wide

Our graphs will be more “quick sketches” than exact graphs.

10

8

6

4

2

-2

-4

-10 -5 5 10

f x = x2

Graph of f(x)=x2

x f(x)

0

1

-1

2

-2

0

1

1

4

4

Axis is x = 0

Vertex at (0, 0)

Points up, opens “normal”

Notice the symmetry

More with Vertex Form

The vertex is (h, k). Changes in (h, k) will shift the quadratic around in the plane (left/right, up/down).

The axis of symmetry is x = h

Notice that you take the opposite of h from how it is written in the

equation

Example #1

Example #2

2( ) ( )f x a x h k

If a > 0, the graph points upIf a < 0, the graph points down

2( ) ( 4)f x x

2( ) 6f x x

Example #32( ) 5( 3) 1f x x

Vertex is _____

Axis is _______

Points _______Vertex is _____Axis is _______Points _______

Vertex is _____Axis is _______Points _______

Vertex is (0, 6)

Axis is x = 0

Points up

Vertex is (-3, -1)

Axis is x = -3

Points up

Vertex is (4, 0)

Axis is x = 4

Points down

Equations of Quadratic Functions

Vertex Form Standard Form

2( ) ( )f x a x h k 2( )f x ax bx c

More with Standard Form

To find the x-value of the vertex, use the formula

To find the y-value, plug in x and solve for y

The axis of symmetry is

Example #1

2( )f x ax bx c

If a > 0, the graph points upIf a < 0, the graph points down

2( ) 4 4f x x x

Vertex is _____

Axis is _______

Points _______

b = 4, a = -1

2

bx

a

2

bx

a

42

2( 1)x

Find x-value of vertex using formula

Find y-value using substitution2( ) 4 3f x x x 2 (2) (2) (2) 1f

Vertex is (2, 1)

Axis is x = 2

Points down

More examplesExample #2

2( ) 3 5f x x x

Vertex is _____

Axis is _______

Points _______

b = -1, a = 3

( 1) 1

2 2(3) 6

bx

a

Find x-value using formula

Find y-value using substitution

2( ) 3( ) 5f x x x 16

1 59

6 12f

Vertex is (1/6, 59/12)

Axis is x = 1/6

Points up

16

16

You try:

Vertex is _____Axis is _______Points _______

2( ) 2 12 1f x x x

Vertex is (3, 17)

Axis is x = 3

Points down

More about a10

8

6

4

2

-5 5

g x = x2

10

8

6

4

2

-5 5

g x = 5x2a =1

a =1/5

a = 5

If a is close to 0, the graph opens _______________If a is farther from 0, the graph opens ____________If a > 0, the graph points________ If a < 0, the graph points ________

10

8

6

4

2

-5 5

g x = 0.2x2

What happens to the graph as the value of a changes?

If a is close to 0, the graph opens widerIf a is farther from 0, the graph opens narrowerIf a > 0, the graph points up If a < 0, the graph points down

When a = 1, the graph is “normal”

Graphing QuadraticsIf you were asked to graph a quadratic,

what information would you need to know to complete the problem?

The vertex, because we need to know where the graph is located in the planeThe value of a, because we need to know if it points up or down, and whether it opens normal, narrow or wide

Our graphs will be more “quick sketches” than exact graphs.

Sketch each quadratic2( ) 5( 3) 1f x x

2( ) 4 3f x x x

2( ) 8 18f x x x

6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

21( ) 4

2f x x

V = (0, 4)Points downWide

V = (-4, 2)Points upNormal

V = (2, 1)Points downNormal

V = (-3, -1)Points upNarrow

Finding x-intercepts of quadratic functions

What are other words for x-intercepts?

Name 4 methods of finding the x-intercepts of quadratic equations:

All are the value of x when y = 0

Summary: Be able to compare and contrast vertex and standard form

Vertex Form Standard Form

How do you find the Vertex?

How do you find the Axis of Symmetry?

How can you tell if the function:• points up or down? •opens normal, wide or narrow?

What info is needed to do a quick sketch or graph?

How do you find the solutions? (x-intercepts, roots, zeroes, value of x when y = 0)

Set = 0, get “squared stuff” alone, then use square root method

Set = 0 and use method of choice (factor, formula or square root)

2( ) ( )f x a x h k 2( )f x ax bx c

Max and Min ProblemsWhat is the definition of the

maximum or minimum point of a quadratic function?

If a quadratic points down, the vertex is a maximum point

The vertex of a quadratic function is either a maximum point or a minimum point

max

min

If a quadratic points up, the vertex is a minimum point

If you are asked to find a maximum or minimum value of a quadratic function, all you need to do is find its vertex

Example

An object is thrown upward from the top of a 100 foot cliff. Its height in feet about the ground after t seconds given by the function f(t) = -16t2 + 8t + 100.

What was the maximum height of the object?

How many seconds did it take for the object to reach its max height?

How can we find the answer? What is the question asking for?

Example

What was the maximum height of the object?

How many seconds did it take for the object to reach its maximum height?

What is the definition of the maximum or minimum point of a quadratic function?

The vertex of a quadratic function is either a maximum point or a minimum point vertex

ExampleStep 1: Visualize the problem

f(t) = -16t2 + 8t + 100.To find the max values, find the vertex

The x-value of the vertex is the max time

The y-value of the vertex is the max height

Output: heightInput: time

8 8 1

2( 16) 32 4

It took about .25 seconds for the object to reach its max height

f(t) = -16t2 + 8t + 100.f(1/4) = -16(1/4)2 + 8(1/4) + 100.

f(1/4) = 101 The max height was 101 feet

(1/4, 101)

Step 2: Understand the equation