Could you graph the rational function     `(2x)/ (x^2+2x-15)`

1 Answer | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`y = (2x)/ (x^2+2x-15)`

 

By factoring `x^2+2x-15` you will get `x^2+2x-15 = (x+5)(x-3)`

 

`y = (2x)/((x+5)(x-3))`

Now you can see that at x = -5 and x = 3 the graph goes to `oo` .

Also we can see that at x = 0 then y = 0. This means the graph goes through origin.

 

`y = (2x)/((x+5)(x-3))`

`y = (2x)/(x^2(1+5/x)(1-3/x))`

`y = 2/(x(1+5/x)(1-3/x))`

 

Now it is clear that when `x rarr+-oo` then `y rarr0` .

`lim_(xrarr+-oo)y = 0`

 

The stationary point of the graph is given by the first derivative.

`y = (2x)/ (x^2+2x-15)`

`dy/dx = ((x^2+2x-15)xx2-2x(2x+2))/(x^2+2x-15)^2`

`dy/dx = (2x^2+4x-30-4x^2-4x)/(x^2+2x-15)^2`

`dy/dx = (-2x^2-30)/(x^2+2x-15)^2`

`dy/dx = -(x^2+15)/(x^2+2x-15)^2`

 

Here `(x^2+15)>0` always. So the graph has no real stationary points.

 

Using the data we can plot the graph.

Sources:

We’ve answered 318,980 questions. We can answer yours, too.

Ask a question