6.6 analyzing graphs of quadratic functions

1. 1
6.6 Analyzing Graphs of 6.6 Analyzing Graphs of
Quadratic FunctionsQuadratic Functions

2. Graphing the parabola Graphing the parabola yy = = f f ((xx) = ) = axax
22
Consider the equation y = x
2
0 1 4 1
(–1, 1) (0, 0) (1, 1) (2, 4)
y
x
4
(–2, 4)
Axis of symmetry: x = 0
( y=x
2
is
symmetric with respect
to the y-axis )
Vertex
(0, 0)
When a > 0, the parabola
opens upwards and is called
concave up.
The vertex is called a
minimum point.
y
210 – 1 – 2 x

3. For the function equation y = x
2
, what is a ?
a = 1 . What if a does not equal 1?
Consider the equation y = – 4x
2
.

0 – 4 – 4
(0, 0)
– 16
y
– 16
(–2, –
16)
(–1, –4)(1, –4)(–4, –16) x
When a < 0, the parabola opens
downward and is called concave down.
The vertex is a maximum point.
y
210 – 1 – 2 x
What is a ?
a = – 4

4. Properties of the Parabola Properties of the Parabola f f ((xx) =) = ax ax
22
The graph of f (x) = ax
2
is a parabola with the
vertex at the origin and the y axis as the line of
symmetry.
If a is positive, the parabola opens upward, if a is
negative, the parabola opens downward.
If |a| is greater than 1 (|a| > 1), the parabola is
narrower then the parabola f (x) = x
2
.
If |a| is between 0 and 1 (0 < |a| < 1), the parabola
is wider than the parabola f (x) = x
2.

5. Algebraic Approach: y = – 4x
2
– 3
-16-40-4-16-4x
2
-19-7-3-7-19-4x
2
-3
210 – 1 – 2 x
Numerical Approach:
Graphical Approach:
Graphing the parabola Graphing the parabola
yy = = f f ((xx) = ) = axax
22
+ + k k
Consider the equation y = – 4x
2
– 3 . What is a ?
a = – 4
x
Vertex
(0, -3)

6. y = – 4x
2
– 3 .
x
Vertex
(0, -3)
y = – 4x
2
In general the graph of y = ax
2
+ k is the graph
of y = ax
2
shifted vertically k units. If k > 0, the
graph is shifted up. If k < 0, the graph is shifted
down. (P. 267)
The graph y = – 4x
2
is shifted
down 3 units.

7. a = – 4. What effect does the 3 have on the
function?
y
x
y= – 4x
2
y= – 4(x – 3)
2
Consider the equation y = – 4(x – 3)
2
. What is a ?
The axis of symmetry is x = 3.
-16-40-4-16 –4 x
2
210 – 1 – 2 x
-4-16-36-64-100-4 (x-3)
2
Numerical Approach:
-36
3
0
Axis of symmetry is shifted 3 units
to the right and becomes x = 3

8. Vertex Form of a Quadratic FunctionVertex Form of a Quadratic Function
The quadratic function
f(x) = a(x – h)
2
+ k, a 0
The graph of f is a parabola .
Axis is the vertical line x = h.
Vertex is the point (h, k).
If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward.
¹

9. 9
6.6 Analyzing Graphs of Quadratic Functions 6.6 Analyzing Graphs of Quadratic Functions
A family of graphs – a group of graphs that displays one or
more similar characteristics.
–Parent graph – y = x
2
Notice that adding a constant to x
2
moves the graph up.
Notice that subtracting a constant from x before squaring it
moves the graph to the right.
y = x
2
y = x
2
+ 2 y = (x – 3)
2

10. 10
Vertex FormVertex Form
Each function we just looked at can be written in the form
(x – h)
2
+ k, where (h , k) is the vertex of the parabola, and
x = h is its axis of symmetry.
(x – h)
2
+ k – vertex form
x = 3(3 , 0)y = (x – 3)
2
or
y = (x – 3)
2
+ 0
x = 0(0 , 2)y = x
2
+ 2 or
y = (x – 0)
2
+ 2
x = 0(0 , 0)y = x
2
or
y = (x – 0)
2
+ 0
Axis of SymmetryVertexEquation

11. 11
Graph TransformationsGraph Transformations
As the values of h and k change, the graph of
y = a(x – h)
2
+ k is the graph of y = x
2

translated.
| h | units left if h is negative, or |h| units right
if h is positive.
| k | units up if k is positive, or | k | units down
if k is negative.

12. 12
Graph a Quadratic Function in Vertex FormGraph a Quadratic Function in Vertex Form
Analyze y = (x + 2)
2
+ 1. Then draw its graph.
This function can be rewritten as y = [x – (-2)]
2
+ 1.
–Then, h = -2 and k = 1
The vertex is at (h , k) = (-2 , 1), the axis of symmetry is x = -2. The
graph is the same shape as the graph of y = x
2
, but is translate 2 units
left and 1 unit up.
Now use the information to draw the graph.
Step 1 Plot the vertex (-2 , 1)
Step 2 Draw the axis of
symmetry, x = -2.
Step 3 Find and plot two
points on one side
of the axis symmetry,
such as (-1, 2) and (0 , 5).
Step 4 Use symmetry to complete the graph.

13. 13
Graph TransformationsGraph Transformations
How does the value of a in the general form
y = a(x – h)
2
+ k affect a parabola? Compare the graphs
of the following functions to the parent function, y = x
2
.
a.
b.
c.
d.
2
2y x=
21
2
y x=
2
2y x=-
21
2
y x=-
y = x
2
a b
dc

14. 14
Example: Write a quadraticExample: Write a quadratic
function for a parabola with a vertex of function for a parabola with a vertex of
(-2,1) that passes through the point (1,-1).(-2,1) that passes through the point (1,-1).
Since you know the vertex, use vertex form!
y=a(x-h)
2
+k
Plug the vertex in for (h,k) and the other point
in for (x,y). Then, solve for a.
-1=a(1-(-2))
2
+1
-1=a(3)
2
+1
-2=9a
a=
-
9
2
1)2(
9
2
2
++
-
= xy
Now plug in a, h, & k!

15. 15
Standard 9 Write a quadratic function in vertex form
Vertex form- Is a way of writing a quadratic equation
that facilitates finding the vertex.
y – k = a(x – h)
2

The h and the k represent the coordinates of
the vertex in the form V(h, k).
The “a” if it is positive it will mean that our
parabola opens upward and if negative it will
open downward.
A small value for a will mean that our
parabola is wider and vice versa.

16. 16
Standard 9 Write a quadratic function in vertex form
Write y = x
2
– 10x + 22 in vertex form.
Then identify the vertex.
y = x
2
– 10x + 22 Write original function.
y + ? = (x
2
 –10x + ? ) + 22Prepare to complete the square.
y + 25 = (x
2
 – 10x + 25) + 22Add
–10
2

2
()=(–5)
2
=25to each side.
y + 25 = (x – 5)
2
+ 22 Write x
2
– 10x + 25 as a
binomial squared.
y + 3 = (x – 5)
2
Write in vertex form.
The vertex form of the function is y + 3 = (x – 5)
2
.
The vertex is (5, –3).
ANSWER

17. 17
EXAMPLE 1 Write a quadratic function in vertex form
Write a quadratic function for the parabola shown.
SOLUTION
Use vertex form because the
vertex is given.
y – k = a(x – h)
2
Vertex form
y = a(x – 1)
2
– 2 Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.
2 = a(3 – 1)
2
– 2 Substitute 3 for x and 2 for y.
2 = 4a – 2 Simplify coefficient of a.
1 = a Solve for a.

18. 18
EXAMPLE 1 Write a quadratic function in vertex form
A quadratic function for the parabola is y = (x – 1)
2
– 2.
ANSWER

19. 19
EXAMPLE 1Graph a quadratic function in vertex form
Graph y –
5
 = – (x + 2)
2
.
1
4
SOLUTION
STEP 1Identify the constants
a = – , h = – 2, and k = 5.
Because a < 0, the parabola
opens down.
1
4
STEP 2Plot the vertex
(h, k) = (– 2, 5) and draw the
axis of symmetry x = – 2.

20. 20
EXAMPLE 1Graph a quadratic function in vertex form
STEP 3 Evaluate the function for two values of x.
x = 0: y = (0 + 2)
2
+ 5 = 4
1
4
–
x = 2: y = (2 + 2)
2
+ 5 = 1
1
4
–
Plot the points (0, 4) and
(2, 1) and their reflections in the
axis of symmetry.
STEP 4Draw a parabola through the
plotted points.

21. 21
GUIDED PRACTICE for Examples 1 and 2
Graph the function. Label the vertex and axis of symmetry.
1. y = (x + 2)
2
– 3 2. y = –(x + 1)
2
+ 5

22. 22
GUIDED PRACTICE for Examples 1 and 2
3. f(x) = (x – 3)
2
– 4
1
2