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