Answer to Question #114632 in Calculus for ANJU JAYACHANDRAN

Question

Trace the curve y=2x^2-8/x^2-16 by showing all the properties used to trace the curve.

Answer 1

1) Domain, vertical asymptotes and intercepts.

The domain is (−∞,−4)∪(−4,4)∪(4,∞), since x cannot be equal to ±4.

Determine the one-sided limits at points -4 and 4 :

$\lim_{ x→−4^+} \frac{2x^2−8}{x^2 −16} =−∞ \\ \lim_{ x→−4^-} \frac{2x^2−8}{x^2 −16} =∞ \\ \lim_{ x→4^+} \frac{2x^2−8}{x^2 −16} =∞ \\ \lim_{ x→4^-} \frac{2x^2−8}{x^2 −16} =−∞$

Therefore, x=−4 and x=4 are the vertical asymptotes.

Determine the x-intercept(s):

$y=0 \\ \frac{2x^2 −8}{x^2-16} =0 \\ \implies 2x^2 −8=0 \implies 2(x^2 −4) = 0 \\ \implies(x−2)(x+2)=0 \implies x=−2, x=2$ .

So, the x-intercepts are (-2,0) and (2,0).

Determine the y-intercept(s):

Now x=0 so y = 1/2, so the y-intercept is (0,1/2).

2) Symmetry

Let y = f(x).

And f(-x)=f(x).

So given curve is y-axis.

3) Monotonicity

$y'=(\frac{2x^2-8}{x^2-16})' = \frac{4x(x^2-16) - 2x(2x^2-8)}{(x^2-16)^2} = \frac{-48x}{(x^2-16)^2} \\ y'=0 \text{ when } x=0$

If x<0, then y>0. If x>0, then y′<0. Therefore, when x<0 the function is increasing, when x>0 the function is decreasing, and x=0 is the maximum.

4) Convex

$y'' = (\frac{2x^2-8}{x^2-16})'' = (\frac{-48x}{(x^2-16)^2})' = \frac{-48(x^2-16)^2 + 48x\cdot2(x^2-16)\cdot(2x)}{(x^2-16)^4} \\ \quad = \frac{48(3x^2+16)}{(x^2-16)^3}$

The second derivative has no roots, but it has the points, when the second derivative does not exist (x=−4,x=4).

If x<−4, then y′′>0. If −4<x<4, then ′′<0. If x>4, then y′′>0. Therefore, when x<−4 and x>4 the function concave upward, and when −4<x<4 the function concave downward.

Draw a graph, given curve is green colour: