Question

the function f is defined as shown.
$f(x) = \frac{1}{x - 3} - \frac{5}{(x - 3)(x + k)}$

• the constant k is a nonzero real number.
• the graph of f has only one vertical asymptote.

what is the positive value of k?
enter your response here:
only 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ..., and / are allowed in your response.
mixed numbers are entered by adding a space after the whole number.
spaces are only allowed between whole numbers and fractions.

Explanation:

Step1: Simplify the function

First, we find a common denominator for the two fractions in \( f(x)=\frac{1}{x - 3}-\frac{5}{(x - 3)(x + k)} \). The common denominator is \( (x - 3)(x + k) \). So we rewrite the first fraction:
\( \frac{1}{x - 3}=\frac{x + k}{(x - 3)(x + k)} \)
Then \( f(x)=\frac{x + k-5}{(x - 3)(x + k)} \)

Step2: Analyze vertical asymptotes

A vertical asymptote occurs where the denominator is zero (and the numerator is not zero at that point). The original denominators are \( x - 3 \) and \( (x - 3)(x + k) \). After simplifying, the denominator of the combined fraction is \( (x - 3)(x + k) \) and the numerator is \( x + k - 5 \).
For the graph to have only one vertical asymptote, the two factors in the denominator must be the same (or one of them makes the numerator zero at the same time, but since \( k eq0 \), we consider the case where the two linear factors are equal or one of them is canceled out by the numerator's root coinciding with a denominator's root, but we want only one vertical asymptote, so we need \( x + k \) to be such that when \( x=-k \), the numerator is also zero (so that the factor \( x + k \) is canceled out, leaving only \( x - 3 \) as the vertical asymptote) or \( x + k=x - 3 \) (but \( k=- 3 \) would make \( x + k=x - 3 \), but let's check the first case).
Set the numerator equal to zero when \( x=-k \):
Substitute \( x=-k \) into the numerator \( x + k-5 \):
\( -k + k-5=-5 eq0 \), so that case doesn't work.
Now, we want the two factors \( x - 3 \) and \( x + k \) to be the same, so \( x + k=x - 3 \) is not possible (since \( k=-3 \), but let's check the simplified function. Wait, maybe we made a mistake. Wait, the original function is \( f(x)=\frac{1}{x - 3}-\frac{5}{(x - 3)(x + k)}=\frac{x + k-5}{(x - 3)(x + k)} \). For there to be only one vertical asymptote, either \( x - 3 \) and \( x + k \) are the same (so \( k=-3 \), but \( k \) is nonzero, but let's check the numerator when \( x + k=x - 3 \), i.e., \( k = - 3 \), then the numerator is \( x-3 - 5=x - 8 \), and the denominator is \( (x - 3)(x - 3)=(x - 3)^2 \). But then the vertical asymptote is at \( x = 3 \), but wait, maybe another approach: if \( x + k \) is such that \( x + k=x - 3 \) is not, but we want only one vertical asymptote, so the root of \( x + k \) must be the same as the root of \( x - 3 \) or the numerator cancels \( x + k \). Wait, no, let's think again. The original function before simplifying has denominators \( x - 3 \) and \( (x - 3)(x + k) \). So the potential vertical asymptotes are at \( x = 3 \) and \( x=-k \). For there to be only one vertical asymptote, either \( x=-k \) is the same as \( x = 3 \) (so \( -k=3\Rightarrow k=-3 \), but \( k \) is nonzero, but let's check the simplified function) or the numerator is zero at \( x=-k \) (so that the factor \( x + k \) is canceled). Wait, when \( x=-k \), numerator is \( -k + k-5=-5 eq0 \), so that can't be. Wait, maybe we made a mistake in simplification. Wait, the first term is \( \frac{1}{x - 3}=\frac{x + k}{(x - 3)(x + k)} \), then subtract \( \frac{5}{(x - 3)(x + k)} \), so \( f(x)=\frac{x + k-5}{(x - 3)(x + k)} \). Now, if \( x + k=x - 3 \), then \( k=-3 \), but then the denominator is \( (x - 3)^2 \) and the numerator is \( x-3 - 5=x - 8 \). Then the vertical asymptote is at \( x = 3 \) (since the numerator is not zero at \( x = 3 \): \( 3-8=-5 eq0 \)). But we need the positive value of \( k \). Wait, maybe we messed up the sign. Wait, maybe the problem is that we want \( x + k \) to be equal to \( x - 3 \) but with a positive \( k \)? No, that would give \( k=-3 \). Wait, maybe another approach: the original function has two potential vertical asymptotes at \( x = 3 \) and \( x=-k \). For there to be only one, one of them must be a hole (i.e., the numerator is zero at that point). So we set the numerator equal to zero at \( x=-k \): \( -k + k-5 = 0 \), but that gives \( -5=0 \), which is impossible. Or set the numerator equal to zero at \( x = 3 \): \( 3 + k-5=0\Rightarrow k = 2 \). Let's check: if \( k = 2 \), then the numerator is \( x+2 - 5=x - 3 \), and the denominator is \( (x - 3)(x + 2) \). So \( f(x)=\frac{x - 3}{(x - 3)(x + 2)}=\frac{1}{x + 2} \) (for \( x eq3 \)). Now, the denominator \( x - 3 \) is canceled by the numerator \( x - 3 \) (except at \( x = 3 \), where the function is undefined, but it's a hole, not a vertical asymptote). So the only vertical asymptote is at \( x=-2 \)? Wait, no, wait: when \( k = 2 \), the original function is \( f(x)=\frac{1}{x - 3}-\frac{5}{(x - 3)(x + 2)}=\frac{x + 2-5}{(x - 3)(x + 2)}=\frac{x - 3}{(x - 3)(x + 2)} \). So we can cancel \( x - 3 \) (for \( x eq3 \)), so the function becomes \( f(x)=\frac{1}{x + 2} \) with a hole at \( x = 3 \). So the vertical asymptote is at \( x=-2 \)? No, that's not right. Wait, no, when we cancel \( x - 3 \), the domain excludes \( x = 3 \) and \( x=-2 \), but the function \( \frac{1}{x + 2} \) has a vertical asymptote at \( x=-2 \), and \( x = 3 \) is a hole (since the limit as \( x\to3 \) of \( f(x) \) is \( \frac{1}{3 + 2}=\frac{1}{5} \), so it's a removable discontinuity (hole) at \( x = 3 \), and a vertical asymptote at \( x=-2 \)? Wait, no, we wanted only one vertical asymptote. Wait, maybe I made a mistake. Let's re - express the function:

\( f(x)=\frac{1}{x - 3}-\frac{5}{(x - 3)(x + k)}=\frac{x + k-5}{(x - 3)(x + k)} \)

We want the graph to have only one vertical asymptote. So either:

1. The factor \( x - 3 \) and \( x + k \) are the same, i.e., \( x + k=x - 3\Rightarrow k=-3 \), but we need positive \( k \), so this is not it.

2. The numerator \( x + k-5 \) has a root that is the same as one of the denominator's roots, so that the corresponding factor is canceled.

Case 1: Root of numerator is \( x = 3 \) (root of \( x - 3 \)):

Set \( 3 + k-5=0\Rightarrow k=2 \)

Case 2: Root of numerator is \( x=-k \) (root of \( x + k \)):

Set \( -k + k-5=0\Rightarrow - 5=0 \), which is impossible.

So when \( k = 2 \), the numerator is \( x+2 - 5=x - 3 \), so \( f(x)=\frac{x - 3}{(x - 3)(x + 2)}=\frac{1}{x + 2} \) (for \( x eq3 \)). The point \( x = 3 \) is a hole (since the limit as \( x\to3 \) is \( \frac{1}{3 + 2}=\frac{1}{5} \)), and the only vertical asymptote is at \( x=-2 \)? Wait, no, that's still one vertical asymptote. Wait, but we wanted the graph to have only one vertical asymptote. Wait, maybe I messed up the initial analysis. Let's check the original function's denominators: \( x - 3 \) and \( (x - 3)(x + k) \). The vertical asymptotes occur where the denominator is zero and the numerator is not zero.

If \( k = 2 \), then the simplified function is \( \frac{1}{x + 2} \) (with a hole at \( x = 3 \)). So the only vertical asymptote is at \( x=-2 \)? No, that's one vertical asymptote. Wait, but we need to find the positive value of \( k \). When \( k = 2 \), we have a hole at \( x = 3 \) and a vertical asymptote at \( x=-2 \)? No, that's not right. Wait, no, when we cancel \( x - 3 \), the function is defined for \( x eq3 \) and \( x eq - 2 \), but the limit as \( x\to3 \) exists (it's \( \frac{1}{5} \)), so \( x = 3 \) is a hole, and \( x=-2 \) is a vertical asymptote. But we wanted only one vertical asymptote. Wait, maybe the other case: if \( x + k=x - 3 \), then \( k=-3 \), but \( k \) is nonzero, but if \( k = 3 \)? Wait, no, let's try \( k = 3 \). Then the function is \( f(x)=\frac{1}{x - 3}-\frac{5}{(x - 3)(x + 3)}=\frac{x + 3-5}{(x - 3)(x + 3)}=\frac{x - 2}{(x - 3)(x + 3)} \). Then we have two vertical asymptotes at \( x = 3 \) and \( x=-3 \), which is not what we want.

Wait, when \( k = 2 \), the numerator is \( x - 3 \), so we can cancel \( x - 3 \) (for \( x eq3 \)), so the function becomes \( \frac{1}{x + 2} \), which has a vertical asymptote at \( x=-2 \) and a hole at \( x = 3 \). So only one vertical asymptote (since the hole is not an asymptote). So that's correct. So the positive value of \( k \) is 2.

Answer:

2