Answer to Question #257667 in Calculus for Pankaj

Question

Trace the curve y= 8/4-x^2 , and state all the properties you use to trace it.

Answer 1

$y=\dfrac{8}{4-x^2}$

1. $4-x^2\not=0=>x\not=\pm2$

Domain: 

$(-\infin;-2)\cup(-2,2)\cup(2, \infin)$

2. $y-$ intersection:

$x=0, y=\dfrac{8}{4-(0)^2}=2$

Point $(0, 2)$

$x-$ intersection(s):

$y=0, \dfrac{8}{4-x^2}=0,$ No solution

There are no $x$ -intersections.

3. $y(-x)=\dfrac{8}{4-(-x)^2}=\dfrac{8}{4-x^2}=y(x)$

The function $y=\dfrac{8}{4-x^2}$ is even on its domain. The garph is symmetric with respect to $y$ -axis.

4.

$\lim\limits_{x\to -2^-}\dfrac{8}{4-x^2}=-\infin$

$\lim\limits_{x\to -2^+}\dfrac{8}{4-x^2}=\infin$

$\lim\limits_{x\to 2^-}\dfrac{8}{4-x^2}=\infin$

$\lim\limits_{x\to 2^+}\dfrac{8}{4-x^2}=-\infin$

Vertical asymptotes: $x=-2, x=2.$

$\lim\limits_{x\to -\infin}\dfrac{8}{4-x^2}=0$

$\lim\limits_{x\to \infin}\dfrac{8}{4-x^2}=0$

Horizontal asymptote: $y=0.$

5.

$y'=(\dfrac{8}{4-x^2})'=\dfrac{16x}{(4-x^2)^2}$

Find the critical number(s):

$y'=0=>\dfrac{16x}{(4-x^2)^2}=0=>x=0, x\not=\pm2$

If $x<-2, y'<0, y$ decreases.

If $-2<x<0, y'<0, y$ decreases.

If $0<x<2, y'>0, y$ increases.

If $x>2, y'>0, y$ increases.

$y(0)=\dfrac{8}{4-(0)^2}=2$

The function $y$ has a local minimum with value of $2$ at $x=0.$

6.

$y''=(\dfrac{16x}{(4-x^2)^2})'=\dfrac{16((4-x^2)^2-2x(4-x^2)(-2x))}{(4-x^2)^4}$

$=\dfrac{16((4-x^2)^2-2x(4-x^2)(-2x))}{(4-x^2)^4}$

$=\dfrac{16(4-x^2+4x^2)}{(4-x^2)^3}=\dfrac{16(4+3x^2)}{(4-x^2)^3}$

If $x<-2, y'<0, y$ is concave down.

If $-2<x<2, y''>0, y$ is concave up.

If $x>2, y''<0, y$ is concave down.

7. Sketch the graph.