Q:

[tex]f(x) = \frac{x^{2} +4x-4}{x^{2} -2x-8}[/tex]Domain:V.A:Roots:Y-int:H.A:Holes:O.A:Also, draw on the graph attached.

Accepted Solution

A:
i) The given function is [tex]f(x)=\frac{x^2+4x-4}{x^2-2x-8}[/tex]We can rewrite in factored form to obtain;[tex]f(x)=\frac{x^2+4x-4}{x^2-2x-8}[/tex][tex]f(x)=\frac{(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)}{(x-4)(x+2)}[/tex]The domain is [tex](x-4)(x+2)\ne0[/tex][tex](x-4)\ne0,(x+2)\ne0[/tex][tex]x\ne4,x\ne-2[/tex]ii) To find the vertical asymptotes equate the denominator to zero.[tex](x-4)(x+2)=0[/tex][tex](x-4)=0,(x+2)=0[/tex][tex]x=4,x=-2[/tex]iii) To find the roots, equate the numerator to zero.[tex](x+2\sqrt{2}+2)(x-2\sqrt{2}+2)=0}[/tex][tex](x+2\sqrt{2}+2)=0,(x-2\sqrt{2}+2)=0}[/tex][tex](x=-2\sqrt{2}-2,x=2\sqrt{2}-2)}[/tex]iv) To find the y-intercept, substitute [tex]x=0[/tex] into the equation.[tex]f(0)=\frac{0^2+4(0)-4}{0^2-2(0)-8}[/tex]We simplify to obtain;[tex]f(0)=\frac{-4}{-8}[/tex][tex]f(0)=\frac{1}{2}[/tex]v) The horizontal asymptote is [tex]lim_{\to \infty}\frac{x^2+4x-4}{x^2-2x-8}=1[/tex]The equation of the horizontal asymptote is y=1vi) The function does not have a variable factor that is common to both the numerator and the denominator.The function has no  holes in it.vii) The given function is a proper rational function.Proper rational functions do not have oblique asymptotes.