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Chapter 1: Problem 16

Find the vertical asymptotes (if any) of the graph of the function. \(h(s)=\frac{2 s-3}{s^{2}-25}\)

Short Answer

Expert verified
The function \(h(s)\) has two vertical asymptotes, located at \(s=5\) and \(s=-5\).

Step by step solution

01

Identify the denominator of the function

The denominator of the function \(h(s)\) is \(s^{2}-25\). The roots of this quadratic expression will give the values of \(s\) for which the function becomes undefined, hence the vertical asymptotes of the function.
02

Set the denominator equal to zero and solve for \(s\)

The equation becomes \(s^{2}-25=0\). This is a difference of squares, which can be factored as \((s-5)(s+5)=0\). Therefore, the values of \(s\) that make this equation true are \(s=5\) and \(s=-5\).
03

Check if these values are excluded values

In some cases, the values that make the denominator equal to zero might not be asymptotes if they also make the numerator equal to zero, hence not being undefined. But in this case, neither \(s=5\) nor \(s=-5\) make the numerator zero. Therefore, both are vertical asymptotes of the function.

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Key Concepts

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

Rational Functions
Rational functions form an important part of algebra, especially when we discuss vertical asymptotes. A rational function is a ratio of two polynomials. Consider the function given in the problem, \( h(s)=\frac{2s-3}{s^{2}-25} \). Here, the numerator is the polynomial \(2s-3\) and the denominator is \(s^{2}-25\). Rational functions can exhibit interesting behaviors like vertical asymptotes, which occur because of the division by zero in the denominator. When dealing with rational functions, it's crucial to identify these behaviors for understanding the function's limits and graphical behavior.
Differences of Squares
The concept of differences of squares is fundamental in solving many polynomial equations, especially when finding vertical asymptotes. A difference of squares is in the format \(a^{2} - b^{2}\), which can be factored as \((a-b)(a+b)\). In our function, the denominator \(s^{2} - 25\) is a difference of squares since it can be rewritten as \(s^{2} - 5^{2}\). Approaching this using the difference of squares method, it factors to \((s-5)(s+5)\). Recognizing and applying this concept helps to quickly find the roots of the denominator, essential for identifying points where the function may be undefined.
Undefined Values
In rational functions, undefined values play a significant role, especially when searching for vertical asymptotes. An undefined value occurs when the denominator of a function is equal to zero, making the function's value undefined. For the function \(h(s)=\frac{2s-3}{s^{2}-25}\), setting the denominator equal to zero \((s^{2} - 25 = 0)\) is crucial. Solving \((s-5)(s+5) = 0\) reveals that the function becomes undefined at \(s = 5\) and \(s = -5\). These undefined points are where we expect to find vertical asymptotes, provided these values do not also zero the numerator.
Factoring Quadratics
Factoring quadratics is a method used to solve equations like the one found in the denominator of a rational function. The quadratic \(s^{2} - 25\) needs to be solved to find where the function becomes undefined. Factoring quadratics involves rewriting the quadratic expression as a product of two binomials. This requires recognizing patterns, such as the difference of squares. Here, \(s^{2} - 25\) factors to \((s-5)(s+5)\). By factoring, we easily find the values \(s = 5\) and \(s = -5\), which are crucial for identifying the points that lead to undefined behavior and potential vertical asymptotes in the graph.

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Most popular questions from this chapter

Use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1]\). Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ f(x)=x^{3}+x-1 $$

At 8:00 A.M. on Saturday a man begins running up the side of a mountain to his weekend campsite (see figure). On Sunday morning at 8:00 A.M. he runs back down the mountain. It takes him 20 minutes to run up, but only 10 minutes to run down. At some point on the way down, he realizes that he passed the same place at exactly the same time on Saturday. Prove that he is correct. [Hint: Let \(s(t)\) and \(r(t)\) be the position functions for the runs up and down, and apply the Intermediate Value Theorem to the function \(f(t)=s(t)-r(t) .]\)

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\left\\{\begin{array}{ll} x, & x \leq 1 \\ x^{2}, & x>1 \end{array}\right. $$

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{3}-x^{2}+x-2, \quad[0,3], \quad f(c)=4 $$

Use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1]\). Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ f(x)=x^{3}+3 x-2 $$
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