Question

How do you graph `y=(x^2-2x)/(2x+3)` using asymptotes, intercepts, end behavior?

Answer 1

graph{(x^2-2x)/(2x+3) [-20, 20, -10, 10]}

Explanation:

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

a)
You can already tell that it has a vertical asymptote. Just let the denominator equal to zero and solve for `x`.
`2x+3=0` so `x=-3/2` (evident on the graph above).

b)
`y` shows zero values when `x=0` and `x=2`

c)
Next, if `x->+-oo`, `y->+-oo`
Roughly speaking;
`y=(oo^2)/(oo)=oo` and `y=oo^2/-oo=-oo`
*Mathematicians hate seeing the infinity symbol used like this. I think it is because infinity cannot simply be seen as a number but more of a concept.

d)
Finally, find the turning points;
`dy/dx=2/(2x+3)^2(x^2+3x-3)=0`
In other words; `x^2+3x-3=0`
Using `(-b+-sqrt(b^2-4ac))/2` you will find `x=(-3+-sqrt(21))/2`.

So far, you have found these important points:
-Asymptote `x=-3/2`
-Points `(0,0)` and `(2,0)`
-`(x->oo,y->oo)` and `(x->-oo,y->-oo)`
-Turning points at `x=(-3+-sqrt(21))/2`

These should be enough to help you see the shape of your graph.
Cheers.