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CHAPTER 3: Quadratic Functions and Equations; Inequalities

3.1 The Complex Numbers3.2 Quadratic Equations, Functions, Zeros, and Models3.3 Analyzing Graphs of Quadratic Functions3.4 Solving Rational Equations and Radical Equations3.5 Solving Equations and Inequalities with Absolute Value

MAT 171 Precalculus AlgebraDr. Claude Moore

Cape Fear Community College

http://cfcc.edu/mathlab/geogebra/quadratic_graph_rootsa.html

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. View PowerPoint presentation

of Quadratic Function.

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.

ExampleFind 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, the x­value at vertex.Minimum value of the function: ­2, the y­value at vertex.

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Example (continued)

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

Example

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

Solution: Complete the square.

Example continued

The graph of g is a vertical shrinking of the graph of y = x2 along with a shift of 4 units to the right.

Graph:

Vertex: (4, 0)Axis of symmetry: x = 4Minimum value of the function: 0

Vertex of a Parabola

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

We calculate the x­coordinate.

We substitute to find the y­coordinate.

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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?

Example (continued)Solutiona) f (x) = ­x2 + 14x ­ 47

The x­coordinate of the vertex is:

Since f (7) = ­(7)2 + 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.c) The range is (−4, 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, ∞).

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?

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.

272/6. Given the quadratic function g(x) = x2 ­ 5x + 6. (a) find the vertex; (b) find the axis of symmetry; (c) determine whether there is a maximum or minimum value and find that value; and (d) graph the function.

272/14. Given the quadratic function f(x) = ­x2 ­ 8x + 5. (a) find the vertex; (b) find the axis of symmetry; (c) determine whether there is a maximum or minimum value and find that value; and (d) graph the function.

(b) axis: x = ­b/(2a) = ­(­8)/(2(­1)) = ­4; x = ­4 yields f(­4) = ­(­4)2 ­ 8(­4) + 5 = 21

(a) vertex: (­4, 21)

(c) Since a = ­1 < 0, the curve opens downward and has a maximum. The maximum value is f(­4) = 21.

273/18. Match the equation y = ­(x ­ 4)2 + 3 with one of the graphs (a) ­ (h).

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273/24. Match the equation y = 2(x ­ 1)2 ­ 4 with one of the graphs (a) ­ (h). 273/28. Determine whether the statement is true or false: The vertex of the graph of the function g(x) = 2(x ­ 4)2 ­ 1 is (­4, ­1).

273/30. Determine whether the statement is true or false: The minimum value of the function f(x) = 3(x ­ 1)2 + 5 is 5.

a) Find the vertex. b) Determine whether there is a maximum or minimum value and find that value. c) Find the range. d) Find the intervals on which the function is increasing and the intervals on which the function is decreasing.273/36. f(x) = ­2x2 ­ 24x ­ 64

a) Find the vertex. b) Determine whether there is a maximum or minimum value and find that value. c) Find the range. d) Find the intervals on which the function is increasing and the intervals on which the function is decreasing.273/40. g(x) = 2x2 ­ 6x + 5

(a) vertex: (1.5, 0.5)

(b) minimum value: y = 0.5

(c) range [0.5, ∞)

(d) increasing to the right of the x­value of the vertex: (1.5, ∞)decreasing to the left of the x-value of the vertex: (-∞, 1.5)

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273/40. Height of a Projectile. A stone is thrown directly upward from a height of 30 ft with an initial velocity of 60. The height of the stone t seconds after it has been thrown is given by the function s(t) = ­16t2 + 60t + 30. Determine the time at which the stone reaches its maximum height and find the maximum height.

274/46. Maximizing Area. A fourth­grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. What is the maximum area that the class can enclose with 32 ft of fence? What should the dimensions of the garden be in order to yield this area?

Since the school serves as one side of the rectangle, they need fence for only three sides: W + L + W = 32 feet of fence. This yields L= 32 ­ 2W.

The area is A = LW = (32 ­ 2W)W = ­2W2 + 32W. Since a = ­2 < 0, this quadratic has a maximum at the vertex. So the dimensions are W = 8 ft and L = 16 ft. Area is A = 128 sq ft.

274/50. Maximizing Profit. In business, profit is the difference between revenue and cost; that is,

Total profit = Total revenue ­ Total cost or P(x) = R(x) ­ C(x),where x is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for

R(x) = 5x and C(x) = 0.001x2 + 1.2x + 60

274/52. Maximizing Profit. In business, profit is the difference between revenue and cost; that is,

Total profit = Total revenue ­ Total cost or P(x) = R(x) ­ C(x),where x is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for

R(x) = 20x ­ 0.1x2 and C(x) = 4x + 2

Sell 80 units for a maximum profit of $638.