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]}

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.

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Answer 2

To graph the function f(x) = x^2/(x+1), we can start by identifying the x and y intercepts.

To find the x-intercept, we set y = 0 and solve for x: 0 = x^2/(x+1) This equation has a hole at x = 0 since the denominator (x+1) cancels out with the numerator (x^2) resulting in 0/1. Therefore, there is no x-intercept.

To find the y-intercept, we set x = 0 and solve for y: y = 0^2/(0+1) y = 0/1 The y-intercept is (0, 0).

Next, let's determine the vertical asymptotes. These occur when the denominator equals zero, so we solve: x + 1 = 0 x = -1 Therefore, there is a vertical asymptote at x = -1.

To find the horizontal asymptote, we examine the behavior of the function as x approaches positive or negative infinity. As x becomes very large or very small, the x^2 term dominates the function. Thus, the horizontal asymptote is y = x^2/x = x.

Now, we can plot the graph using the information we have gathered. We know there is a hole at x = 0, a vertical asymptote at x = -1, and a horizontal asymptote at y = x. The y-intercept is (0, 0).

Please note that the graph may not be accurately represented in this text-based format, so it is recommended to refer to a graphing tool or software for a visual representation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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