dansleiksj

2022-09-30

How do you graph $f\left(x\right)=\frac{{x}^{2}}{x+1}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Expert

You must know that this function means that $y=\frac{{x}^{2}}{x+1}$

To find where x=0, just plug it into the above equation.
$y=\frac{0}{0+1}⇒y=0$

To find where y=0, just plug it into the above equation.
$0=\frac{{x}^{2}}{x+1}⇒x=0$

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.
$\underset{x\to \infty }{lim}\frac{{x}^{2}}{x+1}=\underset{x\to \infty }{lim}\frac{2x}{1}=\infty$
$\underset{x\to -\infty }{lim}\frac{{x}^{2}}{x+1}=\underset{x\to -\infty }{lim}\frac{2x}{1}=-\infty$

So we don't have horizontal asymptotes.

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