# How do you find intercepts, extrema, points of inflections, asymptotes and graph h(x)=xsqrt(9-x^2)?

Aug 22, 2017

#### Explanation:

Domain : $\left[- 3 , 3\right]$

Intercepts
$y$ intercept $h \left(0\right) = 0$

$x$ intercepts: $- 3$, $0$, and $3$
(Solve $h \left(x\right) = 0$)

Asymptotes
The function has no horizontal asymptotes. (The domain is bouinded.) and no vertical asymptotes (there is no division).

$h ' \left(x\right) = \frac{9 - 2 {x}^{2}}{\sqrt{9 - {x}^{2}}}$

Is undefined only at the endpoints of the domain and is $0$ at $x = \pm \frac{3}{\sqrt{2}}$

It is worth noting that as $x$ approaches the endpoints of the domain, $\pm 3$, the derivative goes to $\pm \infty$ So the tangent line becomes vertical.

Local extrema and increasing/decreasing
On $\left[- 3 , - \frac{3}{\sqrt{2}}\right)$, we have $h ' \left(x\right) < 0$ so $h$ is decreasing.
On $\left(- \frac{3}{\sqrt{2}} , 0\right)$, we have $h ' \left(x\right) > 0$ so $h$ is increasing.

And $h \left(- \frac{3}{\sqrt{2}}\right) = - \frac{9}{2}$ is a local minimum.

On $\left(0 , \frac{3}{\sqrt{2}}\right)$, we have $h ' \left(x\right) > 0$ so $h$ is increasing.
On $\left(\frac{3}{\sqrt{2}} , 3\right]$, we have $h ' \left(x\right) > < 0$ so $h$ is decreasing.

And $h \left(\frac{3}{\sqrt{2}}\right) = \frac{9}{2}$ is a local maximum.

Concavity and inflection points

$h ' ' \left(x\right) = \frac{x \left(2 {x}^{2} - 27\right)}{9 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

Is undefined only at the endpoints of the domain and is $0$ at $x = 0$ (The expression is also $0$ at $x = \pm \sqrt{\frac{27}{2}}$, but those are outside the domain of $h$.)

On $\left[- 3 , 0\right)$, we have $h ' ' \left(x\right) > 0$ so the graph of $h$ is concave upwards (convex).

On $\left[- 3 , 0\right)$, we have $h ' ' \left(x\right) < 0$ so the graph of $h$ is concave downwards (concave).

The point $\left(0 , 0\right)$ is the only inflection point.

The graph is:

graph{xsqrt(9-x^2) [-9.23, 13.27, -5.625, 5.625]}