Question

find all vertical asymptotes, ( x = a ), of the following function. for each value of ( a ), evaluate ( lim_{x \to a^+} f(x) ), ( lim_{x \to a^-} f(x) ), and ( lim_{x \to a} f(x) ). use ( infty ) or ( -infty ) when appropriate.

( f(x) = \frac{x^3 - 9x^2 + 14x}{x^2 - 7x} )

...

find all vertical asymptotes. then evaluate ( lim_{x \to a^+} f(x) ), ( lim_{x \to a^-} f(x) ), and ( lim_{x \to a} f(x) ). select the correct choice below, and fill in the answer box if necessary.

\\( \bigcirc \\) a. the vertical asymptote is ( x = \square ). the limits at this vertical asymptote are ( \lim_{x \to a^+} f(x) = \square ), ( \lim_{x \to a^-} f(x) = \square ), and ( \lim_{x \to a} f(x) ) does not exist and is neither ( \infty ) nor ( -\infty ).

\\( \bigcirc \\) b. the vertical asymptote is ( x = \square ). the limits at this vertical asymptote are ( \lim_{x \to a^+} f(x) = \square ), ( \lim_{x \to a^-} f(x) = \square ), and ( \lim_{x \to a} f(x) = \square ).

\\( \bigcirc \\) c. there is no vertical asymptote.

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator:

• Numerator: \(x^3 - 9x^2 + 14x = x(x^2 - 9x + 14) = x(x - 2)(x - 7)\)
• Denominator: \(x^2 - 7x = x(x - 7)\)

Then, cancel out the common factors (note \(x
eq0\) and \(x
eq7\) when simplifying):
\(f(x)=\frac{x(x - 2)(x - 7)}{x(x - 7)} = x - 2\) (for \(x
eq0\) and \(x
eq7\))

Step2: Analyze vertical asymptotes

A vertical asymptote occurs where the function is undefined and the limit is \(\pm\infty\). We check the points where the original function is undefined, i.e., \(x = 0\) and \(x = 7\).

• For \(x = 0\):
• Simplify the limit as \(x\to0\) (using the simplified function, but we need to check the original function's behavior near \(x = 0\)):

\(\lim_{x\to0}\frac{x(x - 2)(x - 7)}{x(x - 7)}=\lim_{x\to0}(x - 2)= - 2\) (the limit exists, so \(x = 0\) is a hole, not a vertical asymptote)

• For \(x = 7\):
• The original function is undefined at \(x = 7\). Now check the left - hand limit (\(x\to7^{-}\)) and right - hand limit (\(x\to7^{+}\)) of the original function (before simplification, but we can also use the fact that near \(x = 7\), \(x

eq7\), so we can use the simplified function's behavior? Wait, no. Wait, when \(x\) approaches \(7\), the original function before cancellation is \(\frac{x(x - 2)(x - 7)}{x(x - 7)}\). If we try to take the limit as \(x\to7\), we can cancel \(x(x - 7)\) only when \(x
eq0\) and \(x
eq7\). So the limit as \(x\to7\) of \(f(x)\) is \(\lim_{x\to7}(x - 2)=7 - 2 = 5\). Wait, but that means \(x = 7\) is also a hole? Wait, no, maybe I made a mistake. Wait, let's re - examine.

Wait, the domain of the original function \(f(x)=\frac{x^3 - 9x^2+14x}{x^2 - 7x}\) is all real numbers except \(x = 0\) and \(x = 7\) (since denominator is zero at \(x = 0\) and \(x = 7\)). Now, for \(x
eq0\) and \(x
eq7\), \(f(x)=x - 2\). So the graph of \(f(x)\) is the line \(y=x - 2\) with holes at \(x = 0\) (where \(f(0)\) would be \(- 2\) if defined) and \(x = 7\) (where \(f(7)\) would be \(5\) if defined). So there are no vertical asymptotes because at the points where the function is undefined (\(x = 0\) and \(x = 7\)), the limit exists (it's equal to the value of the simplified function at those points).

Answer:

C. There is no vertical asymptote