Question

asked Apr 8, 2020 171k views

1 Answer

Jonathan Alfaro · Answer 1 · 2020-04-14T01:57:00+0000

a) The given function is

f(x)=(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

$x^4+x^3-4x^2-4\\e0$

This implies that;

$x\\e -2.69,x\\e 1.83$

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.

(x^2-4)/(x^4+x^3-4x^2-4)=0

$\Rightarrow x^2-4=0$

x^2=4

$x=\pm √(4)$

$x=\pm2$

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)=(0^2-4)/(0^4+0^3-4(0)^2-4)

f(0)=(-4)/(-4)=1

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

$lim_(x\to \infty)(x^2-4)/(x^4+x^3-4x^2-4)=0$

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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