Question

Suppose f(x)=(x+1)/(x^2-4)a) Determine the domain of f(x).
b) Determine the equation of the vertical asymptotes, or state that there are not any.
c) Determine the equation of the horizontal asymptotes, or state that there are not any.
d) Calculate the y-intercept.
e) Calculate the x-intercepts.
f) Locate at least 4 appropriate points and use all of the collected information to sketch a graph.

Answer 1

Answer:

a) Domain of f(x): $x\ \epsilon \ \mathbb{R}-\left \{ -2,2 \right \}$

b) $x=2,x=-2$

c) y=0

d) $\left(0,-\frac{1}{4}\right)$

e) (-1,0)

f) (-4,-0.25) , (-2.2,-1.43) , (-1.8,1.05) , (-1,0) , (0,-0.25) , (1,-0.67) , (1.8,-3.68) , (2.2,3.80) , (4,0.42)

Step-by-step explanation:

$f(x)=\frac{x+1}{x^2-4}$

a) To determinate the domain of f(x), we must analyze if there are possible points of the non-existence of f. They usually come from variable denominators, square roots, logarithms, and trigonometric functions.

In this case, the only possible source of discontinuities come from the denominator. We know the division by 0 is a non valid operation. The domain of f(x) will be the whole set of real numbers except those where

$x^2-4=0$

Factoring

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

Solving

$x=2,x=-2$

Domain of f(x): $x\ \epsilon \ \mathbb{R}-\left \{ -2,2 \right \}$

b) The vertical asymptotes are vertical lines at the zeros of the denominator of a rational function

We already know the donominator has two zeros:

$x=2,x=-2$

Those are the vertical asymptotes

c) To find the horizontal asymptotes we take the limit to infinity of the function:

$\lim\limits_{x \rightarrow \infty}\frac{x+1}{x^2-4}$

Dividing by $x^2$

$\lim\limits_{x \rightarrow \infty}\frac{x/x^2+1/x^2}{x^2/x^2-4/x^2}$

$\lim\limits_{x \rightarrow \infty}\frac{1/x+1/x^2}{1-4/x^2}$

taking the limit, knowing that

$\lim\limits_{x \rightarrow \infty}\frac{k}{x^n}=0 \ (n\geqslant 1)$

$\lim\limits_{x \rightarrow \infty}\frac{1/x+1/x^2}{1-4/x^2}=\frac{0}{1}=0$

The horizontal asymptote is y=0

d) To find the y-intercept, we set x=0

$f(0)=\frac{0+1}{0^2-4}=-\frac{1}{4}$

y-intercept: $\left(0,-\frac{1}{4}\right)$

e) To find the x-intercept, we set y=0 and find the value(s) of x

$\frac{x+1}{x^2-4}=0$ =>

$x+1=0 \ or\ x=-1$

x-intercept: (-1,0)

f) To have a convenient look of the behaviour of f(x) we need at least these set of points

(-4,-0.25) , (-2.2,-1.43) , (-1.8,1.05) , (-1,0) , (0,-0.25) , (1,-0.67) , (1.8,-3.68) , (2.2,3.80) , (4,0.42)

The graph is shown below