Graph each rational function. ƒ(x)=(2x+1)/(x²+6x+8)...

Question

Graph each rational function. ƒ(x)=(2x+1)/(x²+6x+8)

Accepted Answer

Welcome back. I am so glad you're here. We're asked for the following rational function to graph. And our rational function is F of X equals in the numerator five X that's divided by the denominator which is X squared minus X minus 12. And for our answer choices, we have four graphs, we will look at all of the details of those graphs in a little bit. But right now, let's work on building our own graph. Move this to a different part of the screen where there is more space to work and now we'll begin. All right. The first step I like to take is to see if we can factor our function at all and five X in the numerator that's just going to be five X. But looking at the denominator, we can factor X squared minus X minus 12 going to factor this quickly. Well, that'll have a quantity of X minus four, multiplied by the quantity of X plus three. And if that wasn't intuitive, definitely go check out previous lesson videos on factoring. Now that we've got this factored, sometimes we'll use the factored version of the function and sometimes we'll use the original. But beginning now to graph, I'll draw a rough sketch of a coordinate plane. We'll have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle. Now using the six steps from the textbook and from previous lessons, we begin by looking for any vertical as totes and we find the vertical asom totes by setting the denominator equal to zero. And here's where the factored version is very helpful. We can set each of those two factors, our X minus four equal to zero and our X plus three equal to zero. Solving for X adding four to both sides of X minus four equals zero, gives us one vertical Asymptote at X equals a positive four. And solving for X for X plus three equals zero. Subtracting three from both sides. We have another vertical Asymptote at X equals negative three. From the origin for X equals negative three. We'll go to the left three units and draw a vertical asom tote. We can label that X equals negative three. And from the origin for X equals four, we go to the right four units, draw a vertical asom tote and label that X equals four. All right. Step one is done now for step two, step two is to look for horizontal or oblique, add some totes. We do that by comparing the degrees of our numerator and denominator. Here's where the first function is helpful. The degree of our numerator is an unwritten one. The degree of the denominator is two. And when the degree of the numerator is less than the degree of the denominator, by definition, we have a horizontal asom tote at Y equals zero. So we can draw that along the X axis that is now a horizontal Asymptote as well as the X axis double duty here. And so that's Y equals zero as our ASOM tope. Now step three, step three, we're going to see where our Y intercept is. We find the Y intercept by plugging in F of zero by plugging in zero for X. So in our function we have in the numerator five multiplied not by X but by zero all divided by zero squared minus zero minus 12. In the denominator, the numerator simplifies to zero. The denominator simplifies to negative 12 but zero in the numerator tells us that it's going to be zero in the whole thing. So we have a Y intercept when X is zero, Y is zero. That's at the origin, which is also along our horizontal ASOM tote. Interesting. Let's label that 00 for now. And let's keep going. Step four is to find the X intercept or intercepts by setting our numerator equal to zero. That numerator is five, X. Well, dividing both sides by five, we get X equals zero. Well, that matches with the fact that we found a Y intercept at 00 at the origin. So that makes sense so far. But as we said before, that is where we have that horizontal Asymptote. Well, step five is to see if our function intersects the horizontal Asymptote. And we see whether it does intersect the horizontal Asymptote by setting our function equal to the Y intercept, um the Y value of the Y intercept which is zero. So if we set the whole function equal to zero, that will give us where it will intersect with the horizontal asom tote if true, if it makes no sense whatsoever, then it doesn't intersect with the horizontal Asom Tobe. So let's set our function equal to zero. We have five X all divided by, we can use either one this time, we'll use X squared minus X minus 12, all equal to zero to solve. We need to multiply both sides by the denominator. So we're taking zero, multiplied by this X squared minus X minus 12. Well, that's all gone. So when it's multiplied by zero, it's zero. So we just have five X equals zero. Well, we've done this before dividing both sides by five, we get X equals zero. So yeah, it does intersect at the origin. All right. Now step six is that we are going to plot some points in order to do this, we want to choose X values that are in between each of our intervals from negative infinity to the vertical asymptotes and to our X intercept. And in this case, we need an X value from negative infinity to negative three. I chose negative four. You can pick whichever one you want from. We need another X value from negative 3 to 0 where the X intercept is. And I chose negative two. We need another X value from the origin from 00 to our next vertical Asymptote at X equals four. And I chose a positive one and then we need one last X value from the vertical asom tode of X equals four to positive infinity. I chose the X value of five. You're welcome to pause the video plug in those X values into our original function and see what you get for the Y values. I've already calculated this for the X value of negative four. I have negative five halves as the Y value plotting this point from the origin. We're going to the left four units and then we're going down five halves. So that's approximately here for X equals negative two. The Y value is five thirds from the origin. We go to the left two units and then we go up five thirds of a unit. That's approximately here. For the X value of one plugging that into the function, we get a Y value of negative 5/12. So from the origin, we're going to the right one unit and then we're going down 5/12 of a unit, which is approximately here. And then for the point for the X value of five plugging that into the function, we get a Y value of 25 8th. So from the origin, we go to the right five units and then we go up 8th of a unit, which will be approximately here. Now, it's just a matter of drawing the lines for our graph. And so for to the left toward negative infinity for our X values, this graph is going to be hugging along our horizontal asom tote of Y equals zero just below it. And then it will turn and it will go through our point of negative four, negative five halves and then head down toward negative infinity for the Y values along the vertical asom tote of X equals negative three. For the next line, we'll be going to the right of our vertical Asymptote X equals negative three. It will be coming down from positive infinity for the Y values along that asom tote going through a point negative 25 3rd. Then it kind of curves oh curves better than that. It curves toward the origin goes through the origin. That's the place where it goes through the horizontal asym tote goes through our 250. negative five twelveth. And then it will curve down toward negative infinity for the Y values along our vertical Asymptote X equals four. And now to the right of the vertical Asymptote X equals four will be coming down from positive infinity for the Y values along that horizontal Asymptote going through the 40.5 25 8th and then turning and heading toward positive infinity for the X values along our horizontal asom tote of Y equals zero. And there's our graph, we did it. So let's grab this graph shrink it a little bit and see which one it matches given our answer choices. OK. So let's start off looking at asymptotes. We have a horizontal asom tode at Y equals zero. And all four graphs have that great. We have vertical asom totes of X equals negative three and X equals positive four. So answer choice A has vertical asom toads of X equals negative four. That one's ruled out. Answer choice B has a horizon or excuse me, a vertical asom tode of X equals negative four. Nope, not what we're looking for. Answer choice C has horizontal Asymptote. No, a vertical Asymptote. I'm sorry. A vertical Asymptote of X equals negative three and X equals four. That's what we've got in answer choice. D has those vertical asy totes of X equals negative three and X equals four. OK? Answer choice is C and D. What else have you got? Now, let's see where the lines are. We need the one that is most toward negative infinity needs to be down in the third quadrant. And for answer trace C it is down in the third quadrant. But for answer choice D it's up in the second quadrant. And so that one doesn't have the lines in the correct positions from where we found them. So, answer choice, C is our correct answer. Well done. We'll catch you on the next one.

Graph each rational function. ƒ(x)=(2x+1)/(x²+6x+8)

Key Concepts

Rational Functions

Domain and Vertical Asymptotes

Horizontal and Oblique Asymptotes

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Graph each rational function. ƒ(x)=(2x+1)/(x2+6x+8)

Key Concepts

Rational Functions

Domain and Vertical Asymptotes

Horizontal and Oblique Asymptotes

Watch next

Graph each rational function. ƒ(x)=(2x+1)/(x²+6x+8)