1. x-intercepts




That is, when `x=0`


When `x = 0`, `y = 0`.

3. Limit as
x approaches infinity:

Dividing top and bottom by `x^2` gives:

`(9x)/(x^2+9) ÷ x^2/x^2 =(9/x)/(1+9/x^2`

As `x → -∞`,

`y=(9x)/(x^2+9)=(9/x)/(1+9/x^2)` `rarr(0)/(1+0) = 0`

Also, as `x → +∞`, `y → 0`

4. Domain: all real x

Range [see later, in the next section].

5. Maxima and minima?






`(dy)/(dx)=0` when `x=+-3`

So we have a max or min at `(-3, -1.5)` or `(3, 1.5)`.

Some consideration of the expression for `dy/dx` shows that it is

We conclude that `(-3, -1.5)` is the MINIMUM and `(3, 1.5)` is the MAXIMUM.

So the Range is `-1.5 ≤ y ≤ 1.5`.

Also, the slope at `x = 0` is




6. Second derivative:

In this case, the second derivative may be too cumbersome to calculate quickly, and the benefits are doubtful.

So we are ready to sketch the curve:

1020-10-202-2xyOpen image in a new page

Graph of `y=(9x)/(x^2+9)`.

The following features are indicated on the graph:

`x`-intercept `(0,0)` (green dot)

Local maximum `(3,1.5)` (magenta dot)

Local minimum `(-3,1.5)` (magenta dot)

Easy to understand math videos: