### Video Transcript

Which of the following lines is a vertical asymptote of the graph of the function π of π₯ is equal to π₯ cubed minus eight all divided by π₯ squared plus two π₯ minus 15? A) π₯ is equal to five, B) π₯ is equal to two, C) π₯ is equal to eight, or D) π₯ is equal to three.

Weβre asked to check which of the following values of π₯ give vertical asymptotes of the graph of our function π. Well, we can ask the question, what is a vertical asymptote? Well, an asymptote of the graph of the function π is a line which the graph of the function π will approach arbitrarily closely. So a vertical asymptote of the graph of the function π is a vertical line which the graph of the function π approaches arbitrarily closely.

In particular, for rational functions, we know that if π₯ equals π is a vertical asymptote, then the function will not be defined at π. It would be tempting at this point to substitute all of our values of π₯ into our function π. And then call all of the values which give us an undefined value for π of π₯ as vertical asymptotes. If we were to do this, then there would be four possible outputs for our function, since it is a rational function written as a quotient.

We could have that the numerator is zero and the denominator is nonzero. We could have that both the numerator and the denominator are nonzero. We could also have that the numerator is nonzero and the denominator is zero. And the final option we could have is that both the numerator and the denominator are equal to zero.

Firstly, we know that, for any rational function, if π₯ equals π is a vertical asymptote, then the output π of π must be undefined. We see that if our output is of the form of a zero divided by a nonzero, then this will be well defined. We also know that the quotient of two nonzero numbers is also well defined. Therefore, these values cannot be vertical asymptotes in a rational function.

We can therefore ignore these cases since the only time we can get an undefined output in a rational function is when we divide by zero. If the numerator in our output is nonzero and the denominator is zero, we will always have a vertical asymptote at the line π₯ is equal to π. Since our function is rational and written as a quotient. However, when both the numerator and the denominator in our output are equal to zero, we will have a problem.

Consider the rational functions π of π₯ is equal to π₯ minus one divided by π₯ minus one and β of π₯ is equal to π₯ minus one divided by π₯ minus one squared. We see that when we input the value of π₯ equals one into our functions π and β, the output for both is zero divided by zero. However, when we plot these graphs, we see that the graph of π of π₯ is the straight line π¦ is equal to one. With the point at π₯ equals one removed, which we can see graphically has no vertical asymptote when π₯ is equal to one. And we can also see that the graph of β of π₯ is the same as the graph of π¦ equals one over π₯ minus one. Which does have a vertical asymptote when π₯ is equal to one.

So we have a problem when our output is of the form zero over zero, since it could be a vertical asymptote like in β of π₯. But it could also be a removed point like in π of π₯. We know that the only time we will get a zero divided by a zero is when the numerator and the denominator of our function share a factor. If we were to cancel these shared factors, the function will not be changed to any value other than possibly when π₯ is equal to π.

Since weβre only worried about the curve approaching arbitrarily closely to our vertical line π₯ equals π. We do not mind changing the output of our function when π₯ is equal to π. Therefore, we can confidently cancel these shared factors and then reevaluate. We can cancel the shared factors given to us in π of π₯ to just be equal to one. And we can cancel the shared factor of π₯ minus one in β of π₯ to give us one divided by π₯ minus one.

And now since it is clear that π¦ equals one does not have a vertical asymptote when π₯ is equal to one. π of π₯ must also not have a vertical asymptote when π₯ is equal to one. Similarly, for our function π¦ is equal to one over π₯ minus one. When we substitute in the value of π₯ is equal to one, we get a nonzero divided by a zero. So β of π₯ must have a vertical asymptote when π₯ is equal to one.

So letβs clear some space and go back to our question. For our question, we want to find the input values which will make the output of our function undefined. We can do this in two ways. We could substitute the values given to us in the question into our function and then evaluate the outputs. Or we could find the inputs which give us undefined outputs by factoring the denominator of our function.

Since the denominator of our function is a quadratic, we will factor this to find any potential vertical asymptote. We have that π₯ squared plus two π₯ minus 15 can be factored to give π₯ minus three multiplied by π₯ plus five. Since our vertical asymptotes are only possible when the denominator is zero, the only possible vertical asymptotes are the lines π₯ equals three and π₯ is equal to negative five.

We can then substitute the value of π₯ equals three into our function. This gives us an output of 19 divided by zero, which is a nonzero divided by zero. So there must be a vertical asymptote when π₯ is equal to three. We can do the same to evaluate the function when π₯ is equal to negative five. This gives us an output of negative 133 divided by zero.

Since this is of the form of a nonzero number divided by zero, we must have that there is a vertical asymptote when π₯ is equal to negative five. Of these two vertical asymptotes, the only one listed as an option is when π₯ is equal to three. So we can conclude that the graph of the function π has a vertical asymptote when π₯ is equal to three.