Question

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

Answer 1

Please see below.

Explanation:

Domain : `[-3,3]`

Intercepts
`y` intercept `h(0) = 0`

`x` intercepts: `-3`, `0`, and `3`
(Solve `h(x) = 0`)

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

`h'(x) = (9-2x^2)/sqrt(9-x^2)`

Is undefined only at the endpoints of the domain and is `0` at `x = +-3/sqrt2`

It is worth noting that as `x` approaches the endpoints of the domain, `+-3`, the derivative goes to `+-oo` So the tangent line becomes vertical.

Local extrema and increasing/decreasing
On `[-3,-3/sqrt2)`, we have `h'(x) < 0` so `h` is decreasing.
On `(-3/sqrt2,0)`, we have `h'(x) > 0` so `h` is increasing.

And `h(-3/sqrt2) = -9/2` is a local minimum.

On `(0,3/sqrt2)`, we have `h'(x) > 0` so `h` is increasing.
On `(3/sqrt2,3]`, we have `h'(x) ><0` so `h` is decreasing.

And `h(3/sqrt2) = 9/2` is a local maximum.

Concavity and inflection points

`h''(x) = (x(2x^2-27))/(9-x^2)^(3/2)`

Is undefined only at the endpoints of the domain and is `0` at `x = 0` (The expression is also `0` at `x = +-sqrt(27/2)`, but those are outside the domain of `h`.)

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

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

The point `(0,0)` is the only inflection point.

The graph is:

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