Find a rational function ƒ having a graph with the given features...

Question

Find a rational function ƒ having a graph with the given features.
x-intercepts: (1, 0) and (3, 0)
y-intercept: none 
vertical asymptotes: x=0 and x=2
horizontal asymptote: y=1

Accepted Answer

Hey, everyone in this problem, the characteristics of the graph of a rational function F are given below. We're asked to find F of X. We're told that there are two X intercepts at two comma zero and six comma zero that there is no Y intercept that there are vertical asymptotes at X equals zero and X equals five and a horizontal Asymptote at Y equals three. We're given four answer choices A through D option A B and D all have the same numerator which is three multiplied by X minus two, multiplied by X minus six. Option A has the denominator X minus five. Option B has the denominator X and option D has the denominator X multiplied by X minus five. And then answer choice C is F of X is equal to three multiplied by X minus two, divided by X minus five. Let's start with our X intercepts. So we're told that we have X intercepts two comma zero and six comma zero. Now, if you imagine we have a rational function fa numerator divided by some denominator. OK. And we have X intercepts where we set Y equals to zero in order to satisfy that the numerator must be equal to zero. OK? So the numerator must also have X intercepts at these points. OK? So the numerator must have, I'm gonna call them roots of two and six. Now, if a function has roots two and six, we can write these as factors. OK? And we can write the factors as X minus two and X minus six. Yeah, because you can imagine if we have a function factored X minus two, multiplied by X minus six, we set this equal to zero. Then when we solve each factor equal to zero, X minus two equals zero, gives us X equals two and X minus six equals zero, gives us X equals six, which is those X values for those intercepts that we want. OK. So we know that our numerator must have two roots. So we're already looking at two roots in our numerator and so we can eliminate answer choice C OK? It doesn't have the X minus six root in the numerator. And so that's not going to satisfy the conditions we have now having no Y intercept, we're gonna come back to that at the end. Let's try to fill out all of the other information, these conditions that tell us we need to have something and then make sure at the end that we don't have a Y intercept. So looking at our vertical asymptotes, our vertical ascent to recall are going to occur where the denominator is equal to zero. OK. So we saw that when we have X intercepts in a rational function that tells us information about what happens in the numerator, our vertical asymptotes telling us what happens in the denominator. So if we have vertical asymptotes, X equals zero and X equals five, this happens where the denominator is equal to zero. And so the denominator must have roots an X equals zero and X equals five. And again, when I say roots, I mean zeros or X intercepts. Now, if we have roots at X equals zero and X equals five, that means that we have factors of X and X minus five. OK. If we set X equal to zero, well, we get X equal to zero. If we set X minus five equal to zero, we get X is equal to five, which is a second vertical Asymptote. So we need these two factors in the denominator to satisfy both of those vertical assets. So we have two factors in the numerator, two factors in the denominator. What about our horizontal as to, we're told that we have a horizontal ascent to at Y is equal to three. Now, if we have a horizontal Asymptote at Y is equal to something that's not zero, that tells us that the degree of the numerator of our rational function has to equal the degree of the denominator. Now, so far, we found two factors in the numerator, two factors in the denominator. So we have a degree two polynomial in both the numerator and the denominator denominator. Sorry. So, so far that's good. We've satisfied that condition that the degree is equal. OK. How do we get this horizontal as tode of three now? So the degrees are equal and the three comes from the ratio of the leading coefficients. So we want the leading coefficient of the numerator divided by the leading coefficient of the denominator to be equal to three. OK? To get that horizontal as to. So let's try to write out our function. OK. So we have F of X and right now I'm gonna start with just the factors. Hey, we found the numerator must have factors X minus two and X minus six. And the denominator must have factors X and X minus five. So we have X minus two, multiplied by X minus six, divided by X multiplied by X minus five. Now our horizontal ascent to, we need the leading coe of the numerator divided by the leading coefficient of the denominator to be equal to three. How can we do that? Well, it's as simple as putting a constant coefficient of three in front of our numerator. Now, if we were to expand the leading coefficient, the coefficient corresponding with that highest degree term, the X squared term in the numerator would be three. And in the denominator it's just one. So we get three divided by one which gives us three And so we get the horizontal as to what we want. So this function F FX satisfies our vertical asymptotes, our horizontal Asymptote and our X intercepts, we said we would come back to the Y intercepts after we satisfied all of the other conditions and double check that we don't have a horizontal or A Y intercept. Sorry. So A Y intercept is going to occur when X is equal to zero. OK. If we imagine substituting X equals zero in this function, we're gonna be dividing by zero. And so we get no solution there. OK? And so we have no Y intercept, we'll give that a check mark, we've double checked. And so that condition is satisfied as well, which means that this is the final function we were looking for, we found that F of X is equal to three multiplied by X minus two, multiplied by X minus six, divided by X multiplied by X minus five, which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

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Find a rational function ƒ having a graph with the given features. x-intercepts: (1, 0) and (3, 0) y-intercept: none vertical asymptotes: x=0 and x=2 horizontal asymptote: y=1