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Colt1911 [192]
3 years ago
10

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.

Mathematics
1 answer:
Llana [10]3 years ago
6 0

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.

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---

hope it helps

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