Question

title="f(x) = \frac{x^{2} +4x-4}{x^{2} -2x-8}" alt="f(x) = \frac{x^{2} +4x-4}{x^{2} -2x-8}" align="absmiddle" class="latex-formula">
Domain:
V.A:
Roots:
Y-int:
H.A:
Holes:
O.A:

Also, draw on the graph attached.

Answer 1

i) The given function is

$f(x)=\frac{x^2+4x-4}{x^2-2x-8}$

We can rewrite in factored form to obtain;

$f(x)=\frac{x^2+4x-4}{x^2-2x-8}$

$f(x)=\frac{(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)}{(x-4)(x+2)}$

The domain is

$(x-4)(x+2)\ne0$

$(x-4)\ne0,(x+2)\ne0$

$x\ne4,x\ne-2$

ii) To find the vertical asymptotes equate the denominator to zero.

$(x-4)(x+2)=0$

$(x-4)=0,(x+2)=0$

$x=4,x=-2$

iii) To find the roots, equate the numerator to zero.

$(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)=0}$

$(x+2\sqrt{2}+2)=0,(x-2\sqrt{2}+2)=0}$

$(x=-2\sqrt{2}-2,x=2\sqrt{2}-2)}$

iv) To find the y-intercept, substitute $x=0$ into the equation.

$f(0)=\frac{0^2+4(0)-4}{0^2-2(0)-8}$

We simplify to obtain;

$f(0)=\frac{-4}{-8}$

$f(0)=\frac{1}{2}$

v) The horizontal asymptote is

$lim_{\to \infty}\frac{x^2+4x-4}{x^2-2x-8}=1$

The equation of the horizontal asymptote is y=1

vi) 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.