Question

standard 9 homework
score: 6/27 answered: 6/20
question 7
find the vertical asymptote(s) of ( f(x)=\frac{-6 x + 2}{x^{2}-4 x - 5} ).
the vertical asymptote(s) are ( x=)
if there is more than one asymptote, enter your answers separated by a comma.
question help: message instructor post to forum

Explanation:

Step1: Factor the denominator

Factor \(x^{2}-4x - 5\). Using the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\) where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(a = 1\), \(b=-4\), \(c=-5\).
\(x=\frac{4\pm\sqrt{16+20}}{2}=\frac{4\pm\sqrt{36}}{2}=\frac{4\pm6}{2}\). So \(x_1 = 5\), \(x_2=-1\) and \(x^{2}-4x - 5=(x - 5)(x + 1)\).

Step2: Find the vertical asymptotes

The vertical asymptotes of a rational function \(y=\frac{f(x)}{g(x)}\) (where \(f(x)\) and \(g(x)\) are polynomials) occur at the values of \(x\) that make \(g(x)=0\) (provided that \(f(x)
eq0\) at those values).
Set \((x - 5)(x + 1)=0\). Solving \(x-5 = 0\) gives \(x = 5\) and solving \(x + 1=0\) gives \(x=-1\).
Check that \(-6x + 2
eq0\) at \(x = 5\) (\(-6\times5+2=-30 + 2=-28
eq0\)) and at \(x=-1\) (\(-6\times(-1)+2=6 + 2=8
eq0\)).

Answer:

\(-1,5\)