Question

find equations for the vertical asymptotes, if any, for the following rational function.
$f(x) = \frac{14x^2 + 17x - 6}{-7x + 2}$
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Explanation:

Step1: Factor the numerator

First, we factor the numerator \(14x^2 + 17x - 6\). We need two numbers that multiply to \(14\times(-6)= -84\) and add up to \(17\). The numbers are \(21\) and \(-4\). So we can rewrite the middle term:

$$
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$$

So the function becomes \(f(x)=\frac{(7x - 2)(2x + 3)}{-7x + 2}\).

Step2: Simplify the function

Notice that \(-7x + 2=-(7x - 2)\). So we can simplify the function:

$$ f(x)=\frac{(7x - 2)(2x + 3)}{-(7x - 2)}=- (2x + 3)\quad (x eq\frac{2}{7}) $$

This is a linear function (a line) with a hole at \(x = \frac{2}{7}\) (since \(x=\frac{2}{7}\) makes the original denominator zero but we canceled the factor \((7x - 2)\)). A vertical asymptote occurs where the denominator is zero and the numerator is not zero at that point. But since we canceled the common factor, there are no vertical asymptotes (only a hole).

Answer:

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