Question

find equations for the vertical asymptotes, if any, for the following rational function.
$f(x) = \frac{2x^2 + 8x - 10}{-2x + 2}$
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Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(2x^2 + 8x - 10\). We can factor out a 2: \(2(x^2 + 4x - 5)\), then factor the quadratic: \(2(x + 5)(x - 1)\).
The denominator is \(-2x + 2\), factor out a \(-2\): \(-2(x - 1)\).

So the function becomes \(f(x)=\frac{2(x + 5)(x - 1)}{-2(x - 1)}\).

Step2: Simplify (cancel common factors)

Cancel the common factor \((x - 1)\) (note \(x
eq1\) to avoid division by zero). After canceling, we get \(f(x)=-(x + 5)=-x - 5\), which is a linear function (no vertical asymptote, since the original discontinuity at \(x = 1\) is a hole, not a vertical asymptote).

Answer:

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