Question

How do you graph `f(x)=x^2/(x+1)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

graph{x^2 / (x+1) [-10, 10, -5, 5]}

Explanation:

You must know that this function means that `y=x^2/(x+1)`

To find where `x=0`, just plug it into the above equation.
`y=0/(0+1)=>y=0`

To find where `y=0`, just plug it into the above equation.
`0=x^2/(x+1)=>x=0`

To find where we have vertical asymptotes you must find for some "explosive" point. In this case, we have the division by zero (we could have `ln(0)` too, for example).

The point where the function blows up is in `x=-1`. So, here we have a vertical asymptote. Imagine if you plug value near to -1 "to the right". You would have a positive value divided by a very small positive value, it means that the function "goes" to infinity if you come from the right side. If you take very near values by the left you'll have a positive divided by a negative value, so it goes to negative infinity.

Horizontal asymptotes. You must see the limit of the function.
`lim_(x->\infty) x^2 / (x+1) =lim_(x->\infty) (2x)/1 = \infty`
`lim_(x->-\infty) x^2 / (x+1) =lim_(x->-\infty) (2x)/1 = -\infty`

So we don't have horizontal asymptotes.