Answer
A. The domain is $[0, \infty)$
B. The y-intercept is $0$
The x-intercepts are $0$ and $3$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to \infty} (x-3)~\sqrt{x} = \infty$
E. The function is decreasing on the interval $(0,1)$
The function is increasing on the interval $(1, \infty)$
F. The local minimum is $(1, -2)$
G. The graph is concave up on the interval $(0,\infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = (x-3)~\sqrt{x}$
A. The function is defined for all real numbers such that $x \geq 0$.
The domain is $[0, \infty)$
B. When $x = 0$, then $y = (0-3)~\sqrt{0}= 0$
The y-intercept is $0$
When $y = 0$:
$(x-3)~\sqrt{x} = 0$
$x = 0$ or $x=3$
The x-intercepts are $0$ and $3$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to \infty} (x-3)~\sqrt{x} = \infty$
There is no horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \sqrt{x}+(x-3)(\frac{1}{2\sqrt{x}})$
$y' = \frac{(2x)+(x-3)}{2\sqrt{x}}$
$y' = \frac{3x-3}{2\sqrt{x}} = 0$
$3x-3 = 0$
$x = 1$
When $0 \lt x \lt 1$ then $y' \lt 0$
The function is decreasing on the interval $(0,1)$
When $x \gt 1$, then $y' \gt 0$
The function is increasing on the interval $(1, \infty)$
F. When $x = 1$, then $y = (1-3)~\sqrt{1}= -2$
The local minimum is $(1, -2)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(3)(2\sqrt{x})-(3x-3)(\frac{1}{\sqrt{x}})}{4x}$
$y'' = \frac{6x-3x+3}{4x^{3/2}}$
$y'' = \frac{3x+3}{4x^{3/2}} = 0$
$3x+3 = 0$
Since $x=-1$ is not in the domain, there are no values of $x$ in the domain such that $y'' = 0$
When $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the interval $(0,\infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.
