Question

Trace the curve Y=(2x²-8)/(x²-16) by showing all the properties used to trace the curve.

Answer 1

1) Domain, vertical asymptotes and intercepts.

The domain is $(-\infty,-4) \cup (-4, 4) \cup (4, \infty),$ since $x$ cannot be equal to $\pm 4.$

Determine the one-sided limits in points $-4$ and $4.$

$\lim\limits_{x\to -4^{+}}\frac{2x^2-8}{x^2-16} = -\infty \quad \lim\limits_{x\to -4^{-}}\frac{2x^2-8}{x^2-16} = +\infty \\ \lim\limits_{x\to 4^{+}}\frac{2x^2-8}{x^2-16} = +\infty \quad \lim\limits_{x\to 4^{-}}\frac{2x^2-8}{x^2-16} = -\infty$

Therefore, $x=-4 \text{ and } x=4$ are the vertical asymptotes.

Determine the $x\text{-intercept(s)}.$

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

So, the $x\text{-intercepts are } (-2,0) \text{ and } (2,0).$

Determine the $y\text{-intercept(s)}.$

$x=0 \\ y = \frac{2(0)^2 - 8}{(0)^2-16} = \frac{1}{2}$

So, the $y\text{-intercept is } (0,\frac{1}{2}).$

2) Symmetry.

$f(-x) = \frac{2(-x)^2-8}{(-x)^2-16} = \frac{2x^2-8}{x^2-16} = f(x)$

Therefore, the function is even.

3) Non-vertical asymptotes.

Let $y=kx+b$ is an equation of a non-vertical asymptote. Here,

$k = \lim\limits_{x\to\infty}\frac{f(x)}{x} = \lim\limits_{x\to\infty}\frac{2x^2-8}{x(x^2-16)} = 0 \\ b = \lim\limits_{x\to\infty}(f(x) - kx) = \lim\limits_{x\to\infty}\frac{2x^2-8}{x^2-16} = 2$

Therefore, $y=2$ is the horizontal asymptote.

4) 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.

5) 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 $y''<0.$ If $x>4,$ then $y''>0.$ Therefore, when $x<-4 \text{ and } x>4$ the function concave upward, and when $-4<x<4$ the function concave downward.

Draw a graph.