3.3 Analyzing Graphs ofQuadratic Functions

Find the vertex, the axis of symmetry, and the maximum or minimum value of a quadratic function using the method of completing the square.

Graph quadratic functions. Solve applied problems involving maximum and

minimum function values.

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Graphing Quadratic Functions of the Type f (x) = a(x h)2 + k

The graph of a quadratic function is called a parabola.

The point (h, k) at which the graph turns is called the vertex. The maximum or minimum value of f(x) occurs at the vertex. Each graph has a line x = h that is called the axis of symmetry.

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Example

Find the vertex, the axis of symmetry, and the maximum or minimum value of f (x) = x2 + 10x + 23.

Solution: Complete the square.

Vertex: (–5, –2) Axis of symmetry: x = –5

Minimum value of the function: 2

f (x) x2 10x 23

x2 10x 25 25 23

(x2 10x 25) 25 23

(x 5)2 2

x 5 2 2

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Addison Wesley

Example (continued)

Graph f (x) = x2 + 10x + 23

Vertex

2 7

3 2

5 2

7 2

8 7

x y

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Example

Find the vertex, the axis of symmetry, and the

maximum or minimum value of

Solution: Complete the square.

g x x2

2 4x 8.

g x 1

2x2 8x 8

g x 1

2x2 8x 16 16 8

g x 1

2x2 8x 16 1

216 8

g x 1

2x2 8x 16 8 8

g x 1

2x 4 2

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Addison Wesley

Example continued

Graph:

Vertex: (4, 0)

Axis of symmetry: x = 4

Minimum value of the function: 0

g x 1

2x 4 2

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Vertex of a Parabola

The vertex of the graph of f (x) = ax2 + bx + c is

We calculate the x-coordinate.

We substitute to find the y-coordinate.

b

2a, f

b

2a

.

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Example

For the function f(x) = x2 + 14x 47:

a) Find the vertex.

b) Determine whether there is a maximum or minimum value and find that value.

c) Find the range.

d) On what intervals is the function increasing? decreasing?

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Example (continued)

Solution

a) f (x) = x2 + 14x 47

The x-coordinate of the vertex is: b

2a

14

2( 1)7

Since f (7) = 72 + 14 • 7 47 = 2,

the vertex is (7, 2).

b) Since a is negative (a = –1), the graph opens down, so the second coordinate of the vertex, 2, is the maximum value of the function.

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Solution continued

c) The range is (∞, 2].

d) Since the graph opens down, function values increase as we approach the vertex from the left and decrease as we move to the right of the vertex. Thus the function is increasing on the interval (∞, 7) and decreasing on (7, ∞).

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Application - Example

A stonemason has enough stones to enclose a rectangular patio with 60 ft of stone wall. If the house forms one side of the rectangle, what is the maximum area that the mason can enclose? What should the dimensions of the patio be in order to yield this area?

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Example (continued)

1. Familiarize. Make a drawing of the situation, using w to represent the width of the fencing.

2. Translate. Since the area of a rectangle is given by length times width, we have

A(w) = (60 2w)w

= 2w2 + 60w.

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Example (continued)

3. Carry out. We need to find the maximum value of A(w) and find the dimensions for which that maximum occurs. The maximum will occur at the vertex of the parabola, the first coordinate is

Thus, if w = 15 ft, then the length l = 60 2 • 15 = 30 ft and the area is 15 • 30 = 450 ft2.

4. Check. (15 + 15 + 30) = 60 feet of fencing.

5. State. The maximum possible area is 450 ft2 when the patio is 15 feet wide and 30 feet long.

w b

2a

60

2( 2)15.