Question

20%7D" id="TexFormula1" title="f(x) - \frac{x^{2}-4 }{x^{4} +x^{3} -4x^{2}-4 }" alt="f(x) - \frac{x^{2}-4 }{x^{4} +x^{3} -4x^{2}-4 }" align="absmiddle" class="latex-formula">
What is the:

Domain:

V.A:

RootsL

Y-Int:

H.A:

Holes:

O.A:


Also, graph it.

Answer 1

a) The given function is

$f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}$

The domain refers to all values of x for which the function is defined.

The function is defined for

$x^4+x^3-4x^2-4\ne0$

This implies that;

$x\ne -2.69,x\ne 1.83$

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

$x^4+x^3-4x^2-4=0$

This implies that;

$x= -2.69,x=1.83$

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

$\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

$\Rightarrow x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm2$

The roots are $x=-2,x=2$

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

$x=0$ into the function

$f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}$

$f(0)=\frac{-4}{-4}=1$

e) to find the horizontal asypmtote, we take limit to infinity

$lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

The horizontal asymtote is $y=0$

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is  at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.