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7) identify the vertical asymptotes of $f(x) = \\frac{x^2 - 5x + 4}{-4x…

Question

  1. identify the vertical asymptotes of $f(x) = \frac{x^2 - 5x + 4}{-4x - 4}$ teks a2.6(k) a) $x = 4$ b) $x = 1$ c) $x = \frac{1}{4}$ d) $x = -1$

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^2 - 5x + 4\). We need two numbers that multiply to \(4\) and add to \(-5\), which are \(-4\) and \(-1\). So, \(x^2 - 5x + 4=(x - 4)(x - 1)\).

Next, factor the denominator \(-4x - 4\). Factor out \(-4\): \(-4x - 4=-4(x + 1)\).

So the function becomes \(f(x)=\frac{(x - 4)(x - 1)}{-4(x + 1)}\).

Step2: Simplify (if possible)

Check if there are any common factors in the numerator and denominator. The numerator has factors \((x - 4)\) and \((x - 1)\), and the denominator has \(-4\) and \((x + 1)\). There are no common factors, so we proceed to find vertical asymptotes.

Step3: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points, since we already checked for common factors).

Set the denominator equal to zero: \(-4(x + 1)=0\).

Divide both sides by \(-4\): \(x + 1 = 0\).

Solve for \(x\): \(x=-1\).

Now, check the numerator at \(x = -1\): \((-1 - 4)(-1 - 1)=(-5)(-2)=10
eq0\), so there is a vertical asymptote at \(x=-1\).

Answer:

D) \(x = -1\)