Question

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

Answer 1

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

1. $4-x^2 \not =0 \Rightarrow x \not= \plusmn 2$

Domain:

$(−∞;−2)∪(−2,2)∪(2,∞)$

2. y-y− intersection:

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

Point (0, 2)

x− intersection(s):

y=0, $0=\frac{8}{4-x^2}$

No solution

There are no xx -intersections.

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

The function is even on its domain. The garph is symmetric with respect to yy -axis.

4. $lim_{ ​x→−2^- } (\frac{8}{4-x^2}) =−∞$

$lim_{ ​x→−2^+} (\frac{8}{4-x^2}) =∞$

$lim_{ ​x→2^- } (\frac{8}{4-x^2}) =∞$

$lim_{ ​x→2^+ } (\frac{8}{4-x^2}) =−∞$

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

$lim_{ ​x→−∞} (\frac{8}{4-x^2}) =0$

$lim_{ ​x→∞} (\frac{8}{4-x^2}) =0$

Horizontal asymptote: y=0.y=0.

5. $y'°(\frac{8}{4-x^2})'=\frac{16}{(4-x^2)^2}$

Find the critical number(s):

$y'=0 \Rightarrow \frac{16x}{(4-x^2)^2}=0 \Rightarrow x=0, x\not= \plusmn 2$

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.

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

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

6. $y"=(\frac{16x}{(4-x^2)^2})'=\frac{16(4-x^2)^2-2x(4-x^2)(-2x)}{(4-x^2)^4}=\frac{16(4-x^2+4x^2}{(4-x^2)^3}=\frac{16(4+3x^2)}{(4-x^2)^3}$

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.