Precalculus by Richard Wright

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Blessed are the meek, for they will inherit the earth. Matthew‬ ‭5‬:‭5‬ ‭NIV‬‬‬‬‬‬

# 2-02 Quadratic Equations

Summary: In this section, you will:

• Recognize characteristics of parabolas.
• Understand how the graph of a parabola is related to its quadratic function.
• Determine a quadratic function’s minimum or maximum value.
• Solve problems involving a quadratic function’s minimum or maximum value.

SDA NAD Content Standards (2018): PC.4.1, PC.5.3

The water in a fountain follows a curved path. The water is launched from the fountain. Then only gravity influences the water's motion until it hits the pool on the other side. This is called projectile motion and can be modeled by a quadratic function.

A quadratic function is a simple polynomial function where the highest exponent on x is 2. They can be written as f(x) = ax2 + bx + c. The graph of a quadratic function is called a parabola.

## Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola. An important feature of a parabola is its extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. See figure 2.

The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, which are the values of x when y = 0.

#### Example 1: Identify the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in figure 3.

###### Solution

The vertex is the turning points and lowest point on the graph. That appears to be (−2, 0).

The axis of symmetry is the vertical line through the vertex and is x = −2.

The zeros are the x-intercepts where the graph intersects the x-axis. That appears to be (−2, 0).

The y-intercept is where the graph intersects the y-axis. That is (0, 4).

### Quadratic Function and the Characteristics of the Graph

The general form of a quadratic function is

f(x) = ax2 + bx + c

where a, b, and c are real numbers and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward.

The axis of symmetry is found by

$$x = \frac{-b}{2a}$$

Compare that to the quadratic formula.

$$x = \frac{-b±\sqrt{b^2 - 4ac}}{2a}$$

$$= \frac{-b}{2a} ± \frac{\sqrt{b^2 - ac}}{2a}$$

Notice that the first part of the quadratic formula is the axis of symmetry. The second part is the distance the x-intercepts are away from the axis of symmetry.

The y-intercept is easy to find from the general form. Let x = 0 and simplify.

y = ax2 + bx + c

y = a(0)2 + b(0) + c

y = c

The y-intercept is (0, c).

Figure 4 represents the graph of the quadratic function y = x2 + 2x − 3. In this form, a = 1, b = 2, and c = −3. Because a > 0, the parabola opens upward. The axis of symmetry is $$x = \frac{-b}{2a} = \frac{-2}{2⋅1} = -1$$. This makes sense because the vertical line x = −1 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−1, −4). The x-intercepts which are those points where the parabola crosses the x-axis, occur at (−3, 0) and (1, 0).

###### General Form of a Quadratic Function

f(x) = ax2 + bx + c

• Axis of Symmetry: $$x = \frac{-b}{2a}$$
• Vertex: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$
• x-intercepts: $$\left(\frac{-b}{2a} ± \frac{\sqrt{b^2 - ac}}{2a}, 0\right)$$
• y-intercepts: (0, c)

The standard form of a quadratic function is

f(x) = a(xh)2 + k

where (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.

Like in the general form, if a > 0, the parabola opens upward and the vertex is a minimum. If a < 0, the parabola opens downward, and the vertex is a maximum. Figure 5 is the graph of the quadratic function $$y = -\frac{1}{2}(x + 1)^2 + 4$$. Since xh = x + 1 in this example, h = −1. In this example, $$a = -\frac{1}{2}$$, h = −1, and k = 4. Because a < 0, the parabola opens downward. The vertex is at (−1, 4).

The standard form is useful for determining the transformations from the graph of y = x2. Figure 6 is the graph of this basic function.

If k > 0, the graph shifts upward, but if k < 0, the graph shifts downward. In figure 5, k > 0, so the graph is shifted 4 units upward. If h > 0, the graph shifts toward the right and if h < 0, the graph shifts to the left. In figure 5, h < 0, so the graph is shifted 1 units to the left. The magnitude of a indicates the stretch of the graph. If |a| > 1, the graph is stretched vertically, but the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower. But if |a| < 1, the graph is shrunk vertically, and the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider. In figure 5, |a| < 1, so the graph is shrunk vertically and looks wider.

The standard form and the general form are equivalent methods of describing the same function. Explore this by expanding out the general form and setting it equal to the standard form.

a(xh)2 + k = ax2 + bx + c

a(x2 − 2hx + h2) + k = ax2 + bx + c

ax2 − 2ahx + (ah2 + k) = ax2 + bx + c

For the corresponding terms to be equal, the coefficients must be equal. Notice both sets have ax2, so the a's meaning is identical for both forms. Set the middle term's coefficients equal to see how those variables correspond.

−2ah = b

$$h = \frac{-b}{2a}$$

This is the axis of symmetry. In standard form, the axis of symmetry is x = h which is the same as $$x = \frac{-b}{2a}$$ in general form. These are also the x-coordinates of the vertex since it is always on the axis of symmetry.

Set the constant terms equal.

ah2 + k = c

Thus the y-intercept is (0, ah2 + k) in vertex form, or (0, c) in general form.

###### Standard Form of a Quadratic Function

Sometimes called the vertex form.

f(x) = a(xh)2 + k

• Vertex: (h, k)
• Axis of Symmetry: x = h
###### Write the Equation of a Quadratic Function from its Graph
1. Identify the vertex, (h, k).
2. Substitute the values of h and k in the standard form f(x) = a(xh)2 + k
3. Substitute the values of any point on the parabola, other than the vertex, for x and f(x).
4. Solve for the stretch factor, a. (If the parabola opens up, a > 0. If the parabola opens down, a < 0.)
5. Expand and simplify to write in general form.

#### Example 2: Write the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function in figure 7 and simplify to write the equation in general form.

###### Solution

The vertex is (2, −4), so h = 2 and k = −4. Substitute those into the standard form.

f(x) = a(xh)2 + k

f(x) = a(x − 2)2 − 4

Choose some point on the parabola such as (0, −2). Substitute those for x and f(x).

−2 = a(0 − 2)2 − 4

Solve for a.

−2 = 4a − 4

2 = 4a

$$\frac{1}{2} = a$$

Substitute $$a = \frac{1}{2}$$, h = 2, and k = −4 into standard form.

$$f(x) = \frac{1}{2}(x - 2)^2 - 4$$

Expand and simplify.

$$f(x) = \frac{1}{2}(x^2 - 4x + 4) - 4$$

$$f(x) = \frac{1}{2}x^2 - 2x + 2 - 4$$

$$f(x) = \frac{1}{2}x^2 - 2x - 2$$

##### Try It 1

Write an equation for the quadratic function in figure 8 and simplify to write the equation in general form.

$$f(x) = -\frac{1}{2}x^2 + x + \frac{7}{2}$$

#### Example 3: Find the Vertex of a Quadratic Function

Find the vertex of the quadratic function f(x) = 2x2 + 2x − 24. Rewrite the quadratic in standard form.

###### Solution

This is in general form, ax2 + bx + c, so a = 2, b = 2, and c = −24. The x-coordinate of the vertex is at

$$x = \frac{-b}{2a}$$

$$x = \frac{-2}{2⋅2}$$

$$x = -\frac{1}{2}$$

The y-coordinate of the vertex is found by substituting the x-coordinate of the vertex into the function.

f(x) = 2x2 + 2x − 24

$$f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 24$$

$$f\left(-\frac{1}{2}\right) = \frac{1}{2} - 1 - 24$$

$$f\left(-\frac{1}{2}\right) = -\frac{51}{2}$$

The vertex is $$\left(-\frac{1}{2}, -\frac{51}{2}\right)$$

Write the equation in standard form by substituting the vertex (h, k) and a into standard form.

f(x) = a(xh)2 + k

$$f(x) = 2\left(x + \frac{1}{2}\right)^2 - \frac{51}{2}$$

##### Try It 2

Find the vertex and then write the equation in standard form. g(x) = x2 + 2x − 8

(−1, −9); g(x) = (x + 1)2 − 9

## Determine the Maximum and Minimum Values of Quadratic Functions

Many application problems about quadratic functions want the maximum or minimum. These include problems such as, "How high will the ball go?", "What is the maximum revenue?", or "What is the minimum cost?" Because the vertex is the highest or lowest point, the maximum or minimum value is the y-coordinate, or output, of the vertex.

#### Example 4: Find the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a chicken run within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

1. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L.
2. What dimensions should she make her chicken run to maximize the enclosed area?
###### Solution

Use a diagram to visualize the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the chicken run and the length of the fence section parallel to the backyard fence.

1. Area of a rectangle is A = LW. Find an expression for the width by using the part of the perimeter that will be created with the new fence.

P = 2L + W

W = P – 2L

W = 80 – 2L

Substitute this into the area equation.

A = LW

A = L(80 – 2L)

Distribute to put into general form.

A = –2L2 + 80L

2. This is a quadratic function and the maximum will occur at the vertex. Find the L-coordinate (x-coordinate except this function uses L instead of x).

$$x = \frac{-b}{2a}$$

$$L = \frac{-80}{2(-2)} = 20$$

Substitute this into the function to find the area output of the vertex.

A = –2L2 + 80L

A = –2(20)2 + 80(20)

A = 800

The maximum area is 800 ft² and occurs when the sides are 20 ft by 40 ft.

#### Example 5: Find Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local deli currently sells 250 sandwiches during lunch hour at $6 each. When the owner decreased the price by$1, he sold 25 more sandwiches. Assuming that sales are linearly related to the price, what price should the deli charge for a sandwich to maximize their revenue?

###### Solution

Revenue is the amount of money a company brings in. In simple cases like this, the revenue is the product of the price and the sales, R = p⋅s.

If x stands for the number of times the price decreases by $1, then the price is (6 – 1x). Also, the sales would be (250 + 25x). Thus the revenue is R = p⋅s R = (6 – x)(250 + 25x) R = 1500 – 150x – 25x2 R = –25x2 – 150x + 1500 This is a quadratic equation that opens down because a is negative. The maximum will occur at the vertex. The x-coordinate of the vertex is $$x = \frac{-b}{2a}$$ $$x = \frac{150}{2(-25)} = -3$$ This means the price of the sandwiches should decrease –3 times$1, or the price should rise $3. The maximum revenue is found by substituting the x-coordinate of the vertex into the function. R = –25(–3)2 – 150(–3) + 1500 = 2025 The maximum revenue is$2025 when the price is raised $3 so that the deli charges$9 a sandwich. This will of course result in selling 75 fewer sandwiches.

## Find the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will be all y-values greater than or equal to the y-coordinate at the vertex if it opens up, or less than or equal to the y-coordinate at the vertex if the parabola opens down.

###### Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers or (−∞, ∞).

The range of a quadratic function is

• If a > 0, then the range is yf(k) or [k, ∞)
• If a < 0, then the range is yf(k) or (−∞, k]

where k is the y-coordinate of the vertex.

#### Example 6: Find the Domain and Range of a Quadratic Function

Find the domain and range of f(x) = −2x2 − 7x − 4.

###### Solution

For all quadratic function, the domain is all real numbers or (−∞, ∞).

To find the range, start by finding the vertex. The x-coordinate of the vertex is $$x = \frac{-b}{2a}$$.

$$x = \frac{-b}{2a}$$

$$x = \frac{7}{2⋅(-2)} = -\frac{7}{4}$$

Find the y-coordinate of the vertex by substituting the x-coordinate into the function.

f(x) = −2x2 − 7x − 4

$$f\left(-\frac{7}{4}\right) = −2\left(-\frac{7}{4}\right)^2 − 7\left(-\frac{7}{4}\right) − 4$$

$$f\left(-\frac{7}{4}\right) = \frac{17}{8} = k$$

The a of the equation is less than 0, so the parabola opens down and the vertex is the maximum. The range is yk which would be $$y ≤ \frac{17}{8}$$. In interval notation, the range is $$\left(-∞, \frac{17}{8}\right]$$.

##### Try It 3

Find the domain and range of f(x) = x2 + 4x − 21.

Domain: (−∞, ∞); Range: [−25, ∞)

## Find the x- and y-Intercepts of a Quadratic Function

The graph of a quadratic function may have 0, 1, or 2 real x-intercepts. It will always have one y-intercept. The intercepts are useful for applications such as the path of a thrown ball. The x-intercepts would be the places where the ball hits the ground. The y-intercept would be the initial height of the ball.

The x-intercepts are the points where the graph intersects the x-axis, and occur when y = 0. The y-intercept is the point where the graph intersects the y-axis and occurs when x = 0.

###### Find the x- and y-intercepts.
1. Solve the quadratic equation f(x) = 0 to find the x-intercepts.
2. Evaluate f(0) to find the y-intercept.

#### Example 7: Find the x- and y-Intercepts of a Parabola

Find the x- and y-intercepts of f(x) = 2x2 + 5x − 3.

###### Solution

Find the x-intercepts by setting the function equal to zero and solving for x.

2x2 + 5x − 3 = 0

Factor or use the quadratic formula. Factoring is more efficient in this case.

(2x − 1)(x + 3) = 0

Set each factor equal to zero.

2x − 1 = 0    or    x + 3 = 0

2x = 1    or    x = −3

$$x = \frac{1}{2}$$    or    x = −3

The x-intercepts are $$\left(\frac{1}{2}, 0\right)$$ and (−3, 0).

Find the y-intercept by substituting x = 0 into the function.

f(x) = 2x2 + 5x − 3

f(0) = 2(0)2 + 5(0) − 3

f(0) = −3

The y-intercept is (0, −3). See the graph in figure 11.

## Solve Quadratic Equations

In example 7, the quadratic equation used to find the x-intercepts was found by factoring. However, many quadratic equations are not factorable. Two techniques for solving quadratic equations other than factoring are rewriting in standard form and the quadratic formula.

###### Solve a Quadratic Equation by Rewriting in Standard Form
1. Identify a in the general form of the quadratic equation. This is also a in standard form.
2. Substitute a and b into $$x = \frac{-b}{2a}$$ to find the x-coordinate of the vertex.
3. Substitute x = h into the general form of the quadratic function to find k = f(h).
4. Rewrite the quadratic in standard form using a, h, and k.
5. Let the standard form = 0, and solve for x. (These are also the x-intercepts of the graph.)

#### Example 8: Find the x-Intercepts of a Parabola

Find the x-intercepts of the quadratic function f(x) = x2 – 4x – 1.

###### Solution

Find the x-intercepts by setting the function = 0 and solving for x

x2 – 4x – 1 = 0

Unfortunately, this is not easily factored, so rewrite it in standard form. Start by identifying a which is the leading coefficient. In this case a = 1.

Now, find the x-coordinate of the vertex.

$$x = \frac{-b}{2a}$$

$$x = \frac{-(-4)}{2(1)} = 2$$

Because this is the vertex, h = 2. Substitute x = 2 into the function to find the y-coordinate of the vertex.

f(x) = x2 – 4x – 1

f(2) = 22 – 4(2) – 1 = −5

This is the y-coordinate of the vertex, so k = −5. Substitute a, h, and k into standard form.

f(x) = a(xh)2 + k

f(x) = (x − 2)2 – 5

Now find the x-intercepts by setting the function = 0 and solving for x.

0 = (x − 2)2 – 5

5 = (x − 2)2

$$±\sqrt{5} = x - 2$$

$$2 ± \sqrt{5} = x$$

The x-intercepts are $$(2 + \sqrt{5}, 0)$$ and $$(2 - \sqrt{5}, 0)$$.

##### Try It 4

Find the x-intercepts of f(x) = x2 – 6x + 6.

$$(3 – \sqrt{3}, 0)$$ and $$(3 + \sqrt{3}, 0)$$

###### Solve a Quadratic Equation with the Quadratic Formula
1. Write the quadratic in general form = 0, ax2 + bx + c = 0.
2. Identify a, b, and c.
3. Substitute those into the quadratic formula and simplify.

$$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$

#### Example 9: Solve a Quadratic Equation with the Quadratic Formula

Solve 4x2 − 12x − 11 = 0.

###### Solution

Another way to solve quadratic equations is to use the quadratic formula. Start by putting the quadratic into general form. This quadratic already is in general form.

Identify a, b, and c.

ax2 + bx + c = 0

4x2 − 12x − 11 = 0

a = 4, b = −12, and c = −11. Substitute those into the quadratic formula.

$$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$

$$x = \frac{-(-12) ± \sqrt{(-12)^2 - 4(4)(-11)}}{2(4)}$$

$$x = \frac{12 ± \sqrt{320}}{8}$$

$$x = \frac{12 ± 8\sqrt{5}}{8}$$

Factor 4 in the numerator, then reduce the fraction.

$$x = \frac{4\left(3 ± 2\sqrt{5}\right)}{8}$$

$$x = \frac{3 ± 2\sqrt{5}}{2}$$

The solutions are $$x = \frac{3 + 2\sqrt{5}}{2}$$ and $$x = \frac{3 - 2\sqrt{5}}{2}$$.

#### Example 10: Applying the Vertex and x-Intercepts of a Parabola

A baseball is hit straight up at 140 feet per second. The height of the ball when it was hit was 4 feet. The height of the ball above the ground as a function of time can be modeled by h(t) = −16t2 + 140t + 4.

1. When does the ball reach the maximum height?
2. What is the maximum height of the ball?
3. When does the ball hit the ground?
###### Solution
1. The ball reaches the maximum height at the vertex of the parabola. The x-variable in this problem is actually t for time. The question is asking for the time of the maximum height which is the x- (or t-) coordinate of the vertex.

$$t = \frac{-b}{2a}$$

$$t = \frac{-140}{2(-16)}$$

$$t = 4.375$$

The ball reaches the maximum height at 4.375 s after it was hit.

2. The maximum height of the ball is the y- (or h-) coordinate of the vertex. Substitute the t-coordinate of the vertex into the function to the height.

h(t) = −16t2 + 140t + 4

h(4.375) = −16(4.375)2 + 140(4.375) + 4

h(4.375) = 310.25

The maximum height of the ball is 310.25 feet.

3. When the ball hits the ground, its height will be zero. Set the function equal to zero and solve for t.

−16t2 + 140t + 4 = 0

Divide every term by 4 to make all the numbers smaller.

−4t2 + 35t + 1 = 0

This does not look easily factorable, so another method is needed. The quadratic formula is useful with a = −4, b = 35, and c = 1.

$$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$

$$t = \frac{-35 ± \sqrt{(35)^2 - 4(-4)(1)}}{2(-4)}$$

$$t = \frac{-35 ± 35.23}{-8}$$

t = −0.03, 8.78

Traveling back in time is not possible, so only the positive answer will be accepted. The ball hits the ground after 8.78 s.

##### Try It 5

An inept shot putter accidentally threw the ball straight up at 30 feet per second from a height of 7 feet. The ball's height above the ground can be modeled by h(t) = −16t2 + 30t + 7.

1. When does the ball reach the maximum height?
2. What is the maximum height of the ball?
3. When does the ball hit the ground?

a. 0.94 seconds; b. 21.06 feet; c. 2.08 seconds

##### Lesson Summary

###### General Form of a Quadratic Function

f(x) = ax2 + bx + c

• Axis of Symmetry: $$x = \frac{-b}{2a}$$
• Vertex: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$
• x-intercepts: $$\left(\frac{-b}{2a} ± \frac{\sqrt{b^2 - ac}}{2a}, 0\right)$$
• y-intercepts: (0, c)

###### Standard Form of a Quadratic Function

Sometimes called the vertex form.

f(x) = a(xh)2 + k

• Vertex: (h, k)
• Axis of Symmetry: x = h

###### Write the Equation of a Quadratic Function from its Graph
1. Identify the vertex, (h, k).
2. Substitute the values of h and k in the standard form f(x) = a(xh)2 + k
3. Substitute the values of any point on the parabola, other than the vertex, for x and f(x).
4. Solve for the stretch factor, a. (If the parabola opens up, a > 0. If the parabola opens down, a < 0.)
5. Expand and simplify to write in general form.

###### Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers or (−∞, ∞).

The range of a quadratic function is

• If a > 0, then the range is yf(k) or [k, ∞)
• If a < 0, then the range is yf(k) or (−∞, k]

where k is the y-coordinate of the vertex.

###### Find the x- and y-intercepts.
1. Solve the quadratic equation f(x) = 0 to find the x-intercepts.
2. Evaluate f(0) to find the y-intercept.

###### Solve a Quadratic Equation by Rewriting in Standard Form
1. Identify a in the general form of the quadratic equation. This is also a in standard form.
2. Substitute a and b into $$x = \frac{-b}{2a}$$ to find the x-coordinate of the vertex.
3. Substitute x = h into the general form of the quadratic function to find k = f(h).
4. Rewrite the quadratic in standard form using a, h, and k.
5. Let the standard form = 0, and solve for x. (These are also the x-intercepts of the graph.)

###### Solve a Quadratic Equation with the Quadratic Formula
1. Write the quadratic in general form = 0, ax2 + bx + c = 0.
2. Identify a, b, and c.
3. Substitute those into the quadratic formula and simplify.

$$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$

## Practice Exercises

1. What is the advantage of writing a quadratic function in standard form?
2. Rewrite the quadratic functions in standard form and give the vertex.

3. f(x) = x2 – 6x + 10
4. g(x) = 2x2 + 8x + 3
5. h(x) = 3x2 − 24x + 55
6. Determine whether there is a minimum or maximum value for the quadratic function. Then find the value and the axis of symmetry.

7. j(x) = –x2 – 2x – 5
8. k(x) = 3x2 – 54x + 244
9. m(x) = −2x2 – 4x + 8
10. Solve the equation.

11. x2 + 24 = 0
12. x2 − 4x + 1 = 0
13. x2 − 8x + 17 = 0
14. 4x2 − 4x + 13 = 0
15. Find the general form of the equation of the quadratic function shown in the graph.

16. Graph the quadratic function and give the vertex, axis of symmetry, and intercepts.

17. g(x) = −x2 + 2x + 3
18. h(x) = −2x2 − 4x
19. $$j(x) = \frac{1}{2}x^2 + x − 4$$
20. Use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.

21. Vertex (7, 18), opens down
22. Problem Solving

23. Find the dimensions of the rectangular dog run with the greatest enclosed area if there is 50 feet of fencing and one side will be a side of the house and not need fence.
24. A fountain shoots a stream of water from a height of 3 feet at a speed of 24 feet per second. The height in feet of the water can be modeled by h(t) = −16t2 + 24t + 3. What is the maximum height of the water?
25. Coveleski stadium in South Bend, Indiana, holds 5,000 spectators. With a ticket price of $15, the average attendance has been 1,050. When the price dropped to$12, the average attendance rose to 1,200. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?
26. Mixed Review

27. (2-01) Simplify (2 + i)(1 − 2i)
28. (2-01) Simplify $$\frac{2i}{3-i}$$
29. (1-10) The drag, D, of a body falling through a fluid varies directly with the square of the speed, s. If the drag is 90 N at a speed of 30 m/s, find an equation relating D and s. Then what is the drag at 50 m/s?
30. (1-09) Find the inverse of f(x) = x5 − 32
31. (1-05) Find the zeros of k(t) = 2t2t − 1

1. The vertex can be easily identified.
2. f(x) = (x − 3)2 + 1, Vertex (3, 1)
3. g(x) = 2(x + 2)2 − 5, Vertex (−2, −5)
4. h(x) = 3(x − 4)2 + 7, Vertex (4, 7)
5. Maximum is −4 and occurs at (−1, −4); Axis of symmetry is x = −1
6. Minimum is 1 and occurs at (9, 1); Axis of symmetry is x = 9
7. Maximum is 10 and occurs at (−1, 10); Axis of symmetry is x = −1
8. $$-2\sqrt{6} i, 2\sqrt{6} i$$
9. $$2 − \sqrt{3}, 2 + \sqrt{3}$$
10. 4 − i, 4 + i
11. $$\frac{1-2\sqrt{3}i}{2}, \frac{1+2\sqrt{3}i}{2}$$
12. f(x) = −x2 + 2x + 4
13. $$f(x) = \frac{1}{2}x^2 + 2x - 2$$
14. , Vertex: (1, 4), Axis of symmetry: x = 1, Intercepts: (−1, 0), (3, 0), (0, 3)
15. , Vertex: (−1, 2), Axis of symmetry: x = −1, Intercepts: (−2, 0), (0, 0)
16. , Vertex: $$\left(-1, -\frac{9}{2}\right)$$, Axis of symmetry: x = −1, Intercepts: (−4, 0), (2, 0), (0, −4)
17. Domain: (−∞,∞), Range: (−∞, 18].
18. 12.5 feet by 25 feet
19. 12 feet
20. \$18
21. 4 − 3i
22. $$-\frac{1}{5} + \frac{3}{5}i$$
23. $$D = \frac{1}{10}s^2$$; 250 N
24. $$f^{-1}(x) = \sqrt[5]{x+32}$$
25. $$t = -\frac{1}{2}, 1$$