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Chapter 3: Problem 110

Write an equation of a function that meets the given conditions. Answers may vary. \(x\) -intercept: \(\left(\frac{4}{3}, 0\right)\) vertical asymptotes: \(x=-3\) and \(x=-4\) horizontal asymptote: \(y=0\) \(y\) -intercept: (0,-1)

Short Answer

Expert verified
The function is \[ f(x) = \frac{9x - 12}{(x + 3)(x + 4)} \]

Step by step solution

01

- Identify the form of the rational function

Since the horizontal asymptote is at y=0, the degree of the numerator should be less than the degree of the denominator. Assume the function is in the form: \[ f(x) = \frac{a(x - k)}{(x + 3)(x + 4)} \] where a is a constant to be determined, and k is the x-intercept.
02

- Use the given x-intercept

Since the x-intercept is \( \left( \frac{4}{3}, 0 \right) \), substitute \(x = \frac{4}{3} \) into the function and set it equal to 0:\[ 0 = \frac{a(\frac{4}{3} - k)}{((\frac{4}{3} + 3)(\frac{4}{3} + 4))} \]Since the denominator is never 0, the numerator must be zero for \( x = \frac{4}{3} \), hence \( k = \frac{4}{3} \).
03

- Incorporate the x-intercept into the function

Substitute \( k = \frac{4}{3} \) into the rational function:\[ f(x) = \frac{a(x - \frac{4}{3})}{(x + 3)(x + 4)} \]
04

- Use the y-intercept

Since the y-intercept is \( (0, -1) \), substitute \( x = 0 \) and \( f(x) = -1 \) into the function to solve for \(a\):\[ -1 = \frac{a(0 - \frac{4}{3})}{(0 + 3)(0 + 4)} \] \[ -1 = \frac{-\frac{4a}{3}}{12} \] Multiplying both sides by 12 gives: \[ -12 = -\frac{4a}{3} \] Multiplying both sides by 3 gives: \[ -36 = -4a \] Dividing by -4 gives: \[ a = 9 \]
05

- Write the final function

Substitute \( a = 9 \) back into the rational function:\[ f(x) = \frac{9(x - \frac{4}{3})}{(x + 3)(x + 4)} \]Simplifying the numerator gives: \[ f(x) = \frac{9x - 12}{(x + 3)(x + 4)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

x-intercept
The x-intercept is where the graph of the function crosses the x-axis. This happens when y=0. For rational functions, the x-intercept can be found by setting the numerator equal to zero and solving for x.

In our exercise, the x-intercept is given as \(\frac{4}{3}, 0\). This means that when x=\(\frac{4}{3}\), the function equals zero. During the solution process, we incorporated this by placing \(x = \frac{4}{3}\) into the function鈥檚 numerator, ensuring that it equals zero at that point. Thus, we used this information to determine a crucial part of our function鈥檚 structure.
vertical asymptote
Vertical asymptotes are the values of x at which a function approaches infinity or negative infinity. These occur where the denominator of a rational function equals zero and the numerator does not also equal zero.

For our function, the vertical asymptotes are given as x = -3 and x = -4. This tells us that the denominator must have factors of (x + 3) and (x + 4). These make the function undefined at x = -3 and x = -4, causing the function to grow without bound as it approaches these x-values.
horizontal asymptote
Horizontal asymptotes describe the behavior of functions as they approach y as x approaches infinity or negative infinity. For rational functions, these are determined by comparing the degrees of the numerator and the denominator.

In our exercise, the horizontal asymptote is given as y = 0. This condition requires that the degree of the numerator be less than the degree of the denominator. This fits our function since the numerator is a linear polynomial (degree 1) and the denominator is a quadratic polynomial (degree 2).
y-intercept
The y-intercept is where the graph crosses the y-axis, which happens when x=0.

In this exercise, the y-intercept is given as (0, -1), meaning the function equals -1 when x=0. To incorporate this into our function, we substituted x = 0 and set the function equal to -1. This helped us solve for the constant 'a' in our function. By finding 'a', we determined the complete formula of our rational function, ensuring it passes through the given y-intercept.

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Most popular questions from this chapter

Sketch a rational function subject to the given conditions. Answers may vary. Horizontal asymptote: \(y=3\) Vertical asymptotes: \(x=-1\) and \(x=1\) \(y\) -intercept: (0,0) \(x\) -intercept (0,0) Symmetric to the \(y\) -axis Passes through the point (2,4)

A rectangular quilt is to be made so that the length is 1.2 times the width. The quilt must be between \(72 \mathrm{ft}^{2}\) and \(96 \mathrm{ft}^{2}\) to cover the bed. Determine the restrictions on the width so that the dimensions of the quilt will meet the required area. Give exact values and the approximated values to the nearest tenth of a foot.

Determine if the statement is true or false. If 5 is an upper bound for the real zeros of \(f(x)\), then 6 is also an upper bound.

A monthly phone plan costs \(\$ 59.95\) for unlimited texts and 400 min. If more than 400 min are used the plan charges an addition \(\$ 0.45\) per minute. For \(x\) minutes used, the average cost per minute \(\overline{C_{1}}(x)\) (in \$) is given by $$\overline{C_{1}}(x)=\left\\{\begin{array}{ll} \frac{59.95}{x} & \text { for } 0 \leq x \leq 400 \\ \frac{59.95+0.45(x-400)}{x} & \text { for } x>400 \end{array}\right.$$ a. Find the average cost if a customer talks for \(252 \mathrm{~min}\) \(366 \mathrm{~min},\) and \(400 \mathrm{~min} .\) Round to 2 decimal places. b. Find the average cost if a customer talks for \(436 \mathrm{~min}\) \(582 \mathrm{~min},\) and \(700 \mathrm{~min}\). Round to 2 decimal places. c. Suppose a second phone plan costs \(\$ 79.95\) per month for unlimited minutes and unlimited texts. Write an average cost function to represent the average cost \(\overline{C_{2}}(x)\) (in \$) for \(x\) minutes used. d. Find the average cost \(\overline{C_{2}}(x)\) for the second plan for \(252 \mathrm{~min}, 400 \mathrm{~min},\) and \(700 \mathrm{~min}\)

Explain why \(\frac{x^{2}+2}{x^{2}+1}<0\) has no solution.
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