# Video: Graphing a Rational Function Using Derivatives

Which of the following is the graph of the function 𝑓(𝑥) = 1/(10𝑥² + 10𝑥)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

06:27

### Video Transcript

Which of the following is the graph of the function 𝑓 of 𝑥 is one over 10𝑥 squared plus 10𝑥?

We’ve been given a rational function and asked to identify which of the five graphs represents this function. Now there are several things we need to do when looking to identify the graph of a function. Firstly, we look to find any 𝑥- or 𝑦-intercepts. We find the location of any 𝑥-intercepts by setting the function equal to zero and solving for 𝑥. And we find the location of any 𝑦-intercepts by finding the value of the function when 𝑥 is equal to zero.

But let’s have a look at our function. Well, firstly, we should see that there’s no way for our function to be equal to zero. If we have a rational function, it will be equal to zero when the numerator of the function is equal to zero. But the numerator of our function is one. And there’s no way for that to be equal to zero. So there are no 𝑥-intercepts.

Similarly, if we set 𝑥 equal to zero, the denominator of our fraction becomes zero. And we get a function that is of indeterminate form. And we therefore can’t evaluate 𝑓 of zero. There are no 𝑦-intercepts.

Now actually, if we look at all five of our graphs, we see that none of these graphs intersect either the 𝑥- or 𝑦-axes. So what next? Well, next, we consider the location of any asymptotes. Well, vertical asymptotes arise in a number of ways. But primarily, they correspond to the zeros of the denominator of a rational function. So we’re going to set our denominator equal to zero and solve for 𝑥. That is, 10𝑥 squared plus 10𝑥 equals zero. We factor the expression on the left-hand side. And we get 10𝑥 times 𝑥 plus one. And that’s equal to zero.

Now it follows that, for the product of two numbers to be equal to zero, either one or other of those numbers must themselves be equal to zero. So we say either 10𝑥 is equal to zero or 𝑥 plus one is equal to zero. We divide our first equation by 10. And we find 𝑥 is equal to zero. And we subtract one from both sides. And we find our second equation is 𝑥 is equal to negative one. And so we see we have two vertical asymptotes on our graph. They are the lines 𝑥 equals zero and 𝑥 equals negative one.

Now we can actually eliminate one of our graphs. We see graph (D) does indeed have a vertical asymptote at 𝑥 equals zero. But it does not have one at 𝑥 equals negative one. If we look carefully though, we see that (A), (B), (C), and (E) do have vertical asymptotes at 𝑥 equals zero and 𝑥 equals negative one. On graph (B), for example, they’re these two lines.

Next, we’ll look at horizontal asymptotes. Now one way that a horizontal asymptote will occur is if the polynomial in the denominator of the rational function is a higher degree than that of the numerator. In this case, 𝑦 equals zero is a horizontal asymptote. They’ll also occur if both polynomials are the same degree. But actually, in this case, we see that the degree of the polynomial on our denominator is two, whereas on our numerator, it’s zero.

And so we have a horizontal asymptote at 𝑦 equals zero. Now unfortunately, this still doesn’t help us. All of (A), (B), (C), and (E) have that horizontal asymptote. So what next? Well, next, we’re going to consider the existence of critical points of the function.

Critical points — now here we’ll be looking for turning points on our graph — occur when the first derivative of the function is either equal to zero or does not exist. So let’s rewrite our function slightly as 10𝑥 squared plus 10𝑥 to the power of negative one and then differentiate with respect to 𝑥.

Now we’re going to use the general power rule, which is an extension of the chain rule. In this case, we multiply the entire function by the exponent and then reduce that exponent by one. And then we multiply that by the derivative of the inner function. So the derivative of 10𝑥 squared plus 10𝑥, which is 20𝑥 plus 10. That simplifies to negative 20𝑥 plus 10 over 10𝑥 squared plus 10𝑥 all squared.

Now in fact, we can actually divide through by 10. And this simplifies to negative two 𝑥 plus one over 10𝑥 squared times 𝑥 plus one squared. We’re of course interested in the values of 𝑥 such that this is equal to zero. Now for this fraction to be equal to zero, its numerator itself must be equal to zero. So negative two 𝑥 plus one must be equal to zero. And if we divide through by negative one, subtract one, and then divide through by two, we find 𝑥 is equal to negative one-half. So there’s definitely a turning point at 𝑥 equals negative one-half.

Now we might consider where this first derivative does not exist. However, if we do, we find it doesn’t exist at 𝑥 equals zero and 𝑥 equals negative one. We know that there are asymptotes at these points. And so we’re not considering them as turning points.

So let’s look at graphs (A), (B), (C), and (E) and find the ones that have a turning point at 𝑥 equals negative one-half. Well, they are (A) and (E). (B) and (C) appear to have turning points at closer to 𝑥 equals negative 0.75.

We still need to decide between (A) and (E) though. And there are a couple of ways we could do this. We could substitute 𝑥 equals negative one-half into our original function. Notice how in graph (A) the output will be negative and in graph (E) it will be positive.

Alternatively, we could consider the concavity of our curve. We would need to differentiate once again. And when we do, if the second derivative is greater than zero at our turning point, then we know that the graph is concave up there. Whereas if it’s less than zero, then the graph is concave down there. The problem is, our first derivative is in itself quite a nasty rational function. So we would need to use the quotient rule to differentiate once again.

Let’s use the first method. Let’s substitute 𝑥 equals negative one-half into our original function. When we do, we find that 𝑓 of negative one-half is negative two-fifths. We know that for graph (E) we would be expecting a positive output. So we can eliminate that graph, and we see that the answer is graph (A). It’s the graph of the function 𝑓 of 𝑥 is one over 10𝑥 squared plus 10𝑥.