Question

find equations for the vertical asymptotes, if any, for the following rational function.
$f(x) = \frac{-8x^2 + 24x - 16}{-4x + 8}$
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Explanation:

Step1: Simplify the rational function

First, factor the numerator and the denominator.
The numerator: \(-8x^2 + 24x - 16 = -8(x^2 - 3x + 2) = -8(x - 1)(x - 2)\)
The denominator: \(-4x + 8 = -4(x - 2)\)
So, \(f(x)=\frac{-8(x - 1)(x - 2)}{-4(x - 2)}\). We can cancel out the common factor \((x - 2)\) (note that \(x
eq2\) to avoid division by zero in the original function), so \(f(x) = 2(x - 1)=2x - 2\) for \(x
eq2\).

Step2: Analyze vertical asymptotes

A vertical asymptote occurs where the denominator is zero and the numerator is not zero (after simplifying). But here, after canceling the common factor, the denominator of the simplified function has no variable (it's a linear function, not a rational function with a denominator that can be zero for a value of \(x\) that doesn't make the numerator zero). The original denominator is zero at \(x = 2\), but at \(x = 2\), the numerator is also zero (since when \(x = 2\), \(-8(2)^2+24(2)-16=-32 + 48 - 16 = 0\)), so it's a hole (a removable discontinuity) at \(x = 2\), not a vertical asymptote. So there are no vertical asymptotes.

Answer:

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