Find the Holes in the Asymptotes

Question

f(x)= x/25-x^2

I need help determining the holes for this function. They need to be in an ordered pair.

Answer 1

the x values are discontinuous at x =-5 and 5

"holes" in contrast are disconinuities in a line which can be written as an ordered pair (x,y)

the domain discontinuities can be the asymptote type or a line "hole" if its not a vertical line, as that would be a range discontinuity

Answer 2

Hi Meghan,

To start, let's define some important vocabulary. The "holes" you are referring to are more formally known as discontinuities. These occur wherever the function is not defined/not "well behaved" in some way.

There are many different types of discontinuities. For now, all you need to know is that asymptotes represent a type of discontinuity. A vertical asymptote just means that the function approaches infinity or negative infinity as it gets closer to some undefined value of x. If you draw a vertical line at x, it looks like the function approaches that line on either side, but never quite "touches" it.

Returning to your question, another way of asking this is: "For what values of x is the function f(x) = x/25 - x^2 undefined?"

It's slightly unclear whether f(x) is:

Equation 1: (x/25) - x^2

or

Equation 2: x/(25 - x^2)

Equation 1 does not have any "holes" or discontinuities. For all real values of x, you can divide by 25, and then subtract that result by the square of x.

Recall from algebra that a fraction is undefined if the denominator (i.e., the bottom of the fraction) is equal to 0; informally, this is because there is no clear interpretation of how many times 0 can "fit" into a non-zero number. If there is some value of x* where the denominator of a function f(x) is 0, f(x) has a vertical asymptote at x*.

When looking at equation 2, we should ask ourselves: "Are there any values of x where the denominator of the function is equal to 0?" If so, then it does have discontinuities, which are called vertical asymptotes.

We can find these values of x by simply setting the denominator to 0, and solving for x:

(25 - x^2) = 0

This can be done using simple algebra. We can start by isolating the equation so that x is on one side:

25 = x^2

From here, we can take the square root of both sides to get:

x = +5 or -5.

As a quick review: Remember that our solution is both positive and negative 5 as:

5^2 = 5 * 5 = 25

and

(-5)^2 = -5 * -5 = 25.

Thus, for equation 2, there are two vertical asymptotes/discontinuities at x = 5 and x = -5. If you have a graphing calculator, graph the function for yourself to see how this "looks!"

Good luck, and please feel free to reach out if you need any more help.

Nikolai

Answer 3

I assume your equation is x/(25-x²) with the 25 and the squared term in the denominator.

Otherwise there are no asymptotes for (x/25) - x^2

Asymptotes and holes occur when the function has inputs that cause the function to be undefined. In this case, x/(25-x²), if the denominator equals 0, then the function is undefined because division by zero is undefined. So the asymptotes occur when (25-x²)=0, and it can be factored to find its zeros.

I leave the factoring to find the zeros to you, but message if you need more help.

John