standard 9 homework\nscore: 6/27 answered: 6/20\nquestion 7\nfind the vertical asymptote(s) of (…

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

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

Answer

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)\neq0) 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\neq0) at (x = 5) ((-6\times5+2=-30 + 2=-28\neq0)) and at (x=-1) ((-6\times(-1)+2=6 + 2=8\neq0)).

Answer:

(-1,5)