1. x-intercepts

`(9x)/(x^2+9)=0`

when

`9x=0`

That is, when `x=0`

2.
y-intercepts:

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?

`y=(9x)/(x^2+9)`

`(dy)/(dx)=((x^2+9)(9)-(9x)(2x))/((x^2+9)^2)`

`=(-9x^2+81)/(x^2+9)^2`

`=(-9(x^2-9))/((x^2+9)^2)`

Now,

`(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

• POSITIVE for `-3 < x < 3` and
• NEGATIVE for `x < -3` and `x > 3`.

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

`(dy)/(dx)=[(-9x^2+81)/((x^2+9)^2)]_(x=0)`

`=(0+81)/9^2`

`=1`

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:

Open 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:
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