Answer
A. The domain is $(-\infty,-2) \cup (-2,2)\cup (2, \infty)$
B. The y-intercept is $-\frac{1}{4}$
There are no x-intercepts.
C. The function is an even function.
D. $y = 0$ is a horizontal asymptote.
$x = -2$ is a vertical asymptote.
$x = 2$ is a vertical asymptote.
E. The function is decreasing on the intervals $(0,2)\cup (2,\infty)$
The function is increasing on the intervals $(-\infty, -2)\cup (-2,0)$
F. The local maximum is $(0, -\frac{1}{4})$
G. The graph is concave up on the intervals $(-\infty,-2)\cup (2,\infty)$
The graph is concave down on the intervals $(-2,2)$
There are no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{1}{x^2-4}$
A. The function is defined for all real numbers except $x = -2,2$
The domain is $(-\infty,-2) \cup (-2,2)\cup (2, \infty)$
B. When $x = 0$, then $y = \frac{1}{0^2-4} = -\frac{1}{4}$
The y-intercept is $-\frac{1}{4}$
When $y = 0$:
$\frac{1}{x^2-4} = 0$
There are no values of $x$ such that $y = 0$
There are no x-intercepts.
C. $\frac{1}{(-x)^2-4} = \frac{1}{x^2-4}$
The function is an even function.
D. $\lim\limits_{x \to -\infty} (\frac{1}{x^2-4}) = 0$
$\lim\limits_{x \to \infty} (\frac{1}{x^2-4}) = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to -2^-} (\frac{1}{x^2-4}) = \infty$
$\lim\limits_{x \to -2^+} (\frac{1}{x^2-4}) = -\infty$
$x = -2$ is a vertical asymptote.
$\lim\limits_{x \to 2^-} (\frac{1}{x^2-4}) = -\infty$
$\lim\limits_{x \to 2^+} (\frac{1}{x^2-4}) = \infty$
$x = 2$ is a vertical asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{-(1)(2x)}{(x^2-4)^2} = \frac{-2x}{(x^2-4)^2} = 0$
$-2x = 0$
$x = 0$
When $0 \lt x \lt 2$ or $x \gt 2$, then $y' \lt 0$
The function is decreasing on the intervals $(0,2)\cup (2,\infty)$
When $x \lt -2$ or $-2 \lt x \lt 0$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -2)\cup (-2,0)$
F. When $x = 0,$ then $y = \frac{1}{0^2-4} = -\frac{1}{4}$
The local maximum is $(0, -\frac{1}{4})$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(-2)(x^2-4)^2-(-2x)(2)(x^2-4)(2x)}{(x^2-4)^4}$
$y'' = \frac{-2x^2+8+8x^2}{(x^2-4)^3}$
$y'' = \frac{6x^2+8}{(x^2-4)^3} = 0$
$6x^2+8 = 0$
There are no values of $x$ such that $y'' = 0$
When $x \lt -2~~$ or $x \gt 2$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty,-2)\cup (2,\infty)$
When $-2 \lt x \lt 2$, then $y'' \lt 0$
The graph is concave down on the intervals $(-2,2)$
There are no points of inflection.
H. We can see a sketch of the curve below.
