Answer
A. The domain is $(-\infty, -2) \cup (1,\infty)$
B. There is no y-intercept.
The x-intercepts are $-2$ and $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \sqrt{x^2+x-2} = \infty$
$\lim\limits_{x \to \infty} \sqrt{x^2+x-2} = \infty$
There is no horizontal asymptote.
E. The function is decreasing on the interval $(-\infty, -2)$
The function is increasing on the interval $(1, \infty)$
F. There is no local minimum or local maximum.
G. The graph is concave down on the intervals $(-\infty,-2)\cup (1, \infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = \sqrt{x^2+x-2}$
A. The function is defined for all real numbers except when $x^2+x-2 \lt 0$:
$(x+2)(x-1) \lt 0$
$-2 \lt x \lt 1$
The domain is $(-\infty, -2) \cup (1,\infty)$
B. Since $x \neq 0$, there is no y-intercept.
When $y = 0$:
$\sqrt{x^2+x-2} = 0$
$\sqrt{(x+2)(x-1)} = 0$
$x = -2$ or $x=1$
The x-intercepts are $-2$ and $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \sqrt{x^2+x-2} = \infty$
$\lim\limits_{x \to \infty} \sqrt{x^2+x-2} = \infty$
There is no horizontal asymptote.
E. We can try to find the values of $x$ such that $y' = 0$:
$y' = \frac{2x+1}{2~\sqrt{x^2+x-2}} = 0$
Note that $x = -\frac{1}{2}$ is not included in the domain.
There are no values of $x$ in the domain such that $y' = 0$
When $x \lt -2$ then $y' \lt 0$
The function is decreasing on the interval $(-\infty, -2)$
When $x \gt 1$, then $y' \gt 0$
The function is increasing on the interval $(1, \infty)$
F. There is no local minimum or local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(2)(2~\sqrt{x^2+x-2})-(2x+1)(\frac{2x+1}{\sqrt{x^2+x-2}})}{4~(x^2+x-2)}$
$y'' = \frac{(4)(x^2+x-2)-(2x+1)(2x+1)}{4~(x^2+x-2)^{3/2}}$
$y'' = \frac{4x^2+4x-8-4x^2-4x-1}{4~(x^2+x-2)^{3/2}}$
$y'' = \frac{-9}{4~(x^2+x-2)^{3/2}}$
There are no values of $x$ such that $y'' = 0$
When $x \lt -2$ or $x \gt 1$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty,-2)\cup (1, \infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.