1. Analyzing Graphs of Quadratic Functions
Section 3.3
Analyzing Graphs of Quadratic Functions

2. Objectives
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.

3. 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.

4. Example
Find the vertex, the axis of symmetry, and the maximum or minimum value of f (x) = x2 + 10x Then graph the function.
Vertex: (–5, –2) Axis of symmetry: x = –5
Minimum value of the function: 2

5. Example (continued) Graph f (x) = x2 + 10x + 23 x y 5 2 6 1 4 73
Vertex

6. Example
Find the vertex, the axis of symmetry, and the maximum or minimum value of

7. Example continued
Graph: Vertex: (4, 0) Axis of symmetry: x = 4 Minimum value of the function: 0
The graph of g is a vertical shrinking of the graph of y = x2 along with a shift of 4 units to the right.

8. 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.

9. 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?

10. Example (continued)
Solution a) f (x) = x2 + 14x  47 The x-coordinate of the vertex is:
Since f (7) =  • 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.

11. Example 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, ∞).

12. Example
Maximizing Area. A landscaper has enough stone to enclose a rectangular koi pond next to an existing garden wall of the Englemans’ house with 24 ft of stone wall. If the garden wall forms one side of the rectangle, what is the maximum area that the landscaper can enclose? What dimension of the koi pond will yield this area?

13. Example (continued)
1. Familiarize. Make a drawing of the situation, using w to represent the width of the koi pond, in feet.
2. Translate. Since the area of a rectangle is given by length times width, we have
A(w) = (24  2w)w
=  2w2 + 24w.

14. 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 = 6 ft, then the length l = 24  2 • 6 = 12 ft and the area is 12 • 6 = 72 ft2.

15. Example (continued)
4. Check. As a partial check, we note that 72 ft2 > 40 ft2, whish is larger than the area found in Step 1. As a more complete check, we could examine a table of values for A(w) = (24 – 2w)w and/or examine its graph. 5. State. The maximum possible area is 72 ft2 when the koi pond is 6 ft wide and 12 ft long.