In Exercises 81–88, a. Find the slant asymptote of the graph o...

Question

In Exercises 81–88, a. Find the slant asymptote of the graph of each rational function and b. Follow the seven-step strategy and use the slant asymptote to graph each rational function. f(x)=(x³+1)/(x²+2x)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to determine the equation of the Slant Asom tote of the graph of the given rational function. And to graph the given rational function following the seven steps, our given rational function is F of X equals in the numerator. We have the quantity of X cubed plus five. And that's divided by our denominator which is two X squared plus X. And we have for answer choice A, a Slant Asom tote at Y equals one half X minus 1/4. And then there's a graph for answer choice B. Slant Asimo is Y equals one half X minus 1/4. And then there is a graph for answer choice C A slant Asom tote of Y equals negative one half X plus 1/4. And then we have a graph and answer choice D A Slant Asom tote at Y equals negative one half X minus 1/4. And then we have a graph. All right, we will evaluate these graphs after completing the seven steps for graphing them first though we are going to determine the Slant Asom tote. So to determine whether we're, we have a S slant. Asom tote, we look at the degree of the numerator and the degree of the denominator. The degree of our numerator is three. The degree of our denominator is two. If the degree of the numerator is one more than the degree of the denominator, we do have a slant Asymptote and in this case, it is three is one greater than two. So we do have a slant asom tote now to determine what the slant Asymptote is, we're going to divide. So we'll take our numerator which is X cubed plus five, but we want to enumerate each degree. So we have X cubed plus, we have zero X squared, then plus zero, excuse me, zero X is to the first degree and then plus five, we're dividing that by our denominator which is two X squared plus X and sometimes doing division like this is great and sometimes it's like this. So what we do is we ask ourselves how many times does two X squared go into X cubed? Or in other words, we're taking X cubed and dividing it by two X squared. Well, both of those are divisible by X squared. So if we divide X cubed by X squared, that leaves with one X in the numerator that's divided by two X squared, divided by X squared is just two. So we have X divided by two, that's to the first degree. So we put that over our zero X So we have X divided by two. Now we take that X divided by two and multiply it by both two X squared and positive X taking our X divided by two, multiplied by the two X squared. That's going to get us back to our X cubes that we had. And for X divided by two, multiplied by X, that's a positive X squared divided by two. Now we're going to take X cubed plus zero X squared and subtract the quantity of X cubed plus X squared divided by two. Or in other words, be sure to subtract both of those terms. Well, X cubed minus X cubed is no more X cubed and zero X squared minus X squared divided by two is a negative X squared divided by two. Now we can bring down this plus zero X. The next bit of fun is we ask ourselves how many times does two X squared go into a negative X squared divided by two? So this time we're taking negative X squared divided by two divided by two X squared, which is the same as negative X squared divided by two multiplied by the reciprocal of two X squared, which is one divided by two X squared. Now those X squared are going to cancel out. So we'll just have in the numerator a negative one that's divided by two multiplied by two, which is four. So this is a negative 1/4. So we have XX divided by two minus 1/4 and we could keep going with this division. However, thankfully, for Slant Asom totes, we don't need to worry about the remainder. This is all we need. So we've reached the end of the line. We've gotten our X to the first degree in our constant terms. So our slant Asom tote is Y equals and then what we've solved here and we have X divided by two, but we can factor out in front of that. That's the same as one half X and then minus the 1/4 that's the same thing. So we've solved the first part, which is great. Now, we just have to complete all seven steps in order to graph this. So still a bit more work that we need to accomplish. So I am going to move our screen a little bit so that we can begin to graph to graph. I'm going to draw a rough sketch of a coordinate plane. We have a vertical Y axis, a horizontal X axis, they come together at the origin in the middle. Now, the first step in our seven step process of graphing is we want to determine if the graph has symmetry and we do that by plugging in for our X, we're going to plug in F of negative X and see what we get. So if we plug it in a negative X in the numerator, we'll have a negative X cubed plus five. That's all divided by, in the denominator two multiplied by negative X squared plus a negative X. Now simplifying in the numerator, negative X cubed is now negative X then plus five all divided by negative X squared is a positive X squared. So we've got two X squared or excuse me in the numerator that's still cubed still, but it's negative. And then this negative X is now a negative. So where have the signs changed? Where have they remained the same? Well, the sign changed for our negative X cubed in the numerator and for the negative X to the first degree in the denominator. However, the signs, the sign did not change for our two X squared. So in order to have symmetry, one of two things needs to happen, we need to either come out with everything being the opposite or negative F of X. If we had that, that would indicate symmetry about the origin. But we don't have that. We had two signs change. One didn't and I mean, the fives sign didn't change either. And the other possibility would need to be for symmetry that we would get our original F of X that nothing would change. But we don't have that. We had two things change. So there is no symmetry, sorry. Symmetry. You do not exist for this one. So no, that's our step one. Got six more to go. Well, we can do this, we've got this, some of them are shorter than others. So step two, we are going to determine if we have any Y intercepts to determine if we have Y intercepts. We are going to evaluate for F of zero plugging zero in for X. So in the numerator, we have zero cubed plus five. That's all divided by two, multiplied by zero squared plus not X but zero in the numerator zero cubed plus five, that's five. But in the denominator zero squared is zero, multiplied by two is zero plus zero is zero. Ah We've got a zero in the denominator that's undefined. We cannot have that. There are no Y intercepts not allowed here for this problem. Now, the next thing we do are step three is to determine our X intercepts to determine the X intercepts. We are going to set our numerator equal to zero X cubed plus five. And if you need any refreshers on any of these steps, please check out previous lesson videos on these seven steps. So for X intercept equals or for our X intercepts, our numerator equals zero, we can subtract five from both sides. Then we have X cubed equals and negative five. We'll take the cubed root of both sides. And on the left, we just have X ya on the right, that's equal to the cubed rut of negative five plugging that into a calculator. That is approximately negative 1.71. So we have an X intercept at negative 1. zero. All right, we've got something, we've got an X intercept. Very exciting. Now, step four of our seven step process is we are going to determine the vertical asom totes. So for our vertical asom totes, we are going to set the denominator equal to zero. Our denominator is two X squared plus X set that equal to zero. We can factor out an X. So we'll have X multiplied by the quantity of two X plus one, all equals zero. Now we set both of those two factors equal to zero. So we have an X equals zero. And then our second factor two X plus one equals zero still have to solve for X. On that one, we'll subtract one from both sides. We have two X equals negative one, divide both sides by two and we get X equals a negative one half. So we have two vertical asom totes one at X equals +01 at X equals negative one half. And we have completed step four. We're over halfway done. That's exciting. And step five is that we are going to determine any horizontal ascent totes. This is similar to what we started doing actually for horizontal asymptotes, we look at the degree of the numerator and the degree of the denominator. Here, the numerator's degree is one more than the degree of the denominator. When the degree of the numerator is bigger, then we don't have any horizontal asom totes. But when the degree of the numerator is one larger than the denominator, we do have the slant asom tote, which we've already solved for. And so we've gotten that step taken care of two more. We're getting there. Step six is to plot some points. We need to plot the points between all of our vertical asom totes and our X intercept. So now I'm going to sort of begin doing step seven, step seven is going to be to graph it. So we'll graph some of it so that it's easier to plot the points. So let's graph our vertical asom totes, we have one at X equals zero. That's along the Y axis. So we can draw that asom tote along the Y axis that's at X equals zero. We have another vertical asym tode at X equals negative one half. So from the origin, going to the left one half of a unit, it's going to be very zoomed in for our X axis so that we can see what's going on. So that's X equals a negative one half. And then our X intercept that was at negative 1.710. So from the origin going to the left 1.71 units, so we can label this, it's approximately here at negative 1.710. And then we can draw our slant asom tote which is at Y equals one half, X minus 1/4. So we note that our slope is positive. And our Y intercept for this asom tote is at negative 1/4. So we can draw it. It's gonna look approximately like this. That's our slant asom tote of Y equals one half X minus 1/4. So it's a very, very small slope. And now we need to choose our X values to plot our points. We need to choose X values for our, each of our four intervals. The first interval is from negative infinity to negative 1.71. Our second interval is from negative 1.71 to negative one half. Our third interval is from negative one half to zero. And our fourth interval is from zero to positive infinity. So we will see what's going on and you do not have to choose these X values. But the ones I chose are for the first interval, negative two, for the second interval, negative one. For the third interval, I chose negative 1/4 and for the fourth interval, I chose a positive one. Now you're welcome to pause the video and plug those X values into our X cubed plus five divided by two X squared plus X function. But I am going to continue. And our value for Y when X equals negative two is negative one half. Our value for Y when X equals negative one is a positive four. Our value for Y when X equals negative 1/4 is approximately negative 39. and our Y value when X equals one is a positive two. So now we can plot those. So from the origin for our first point negative two, negative one half, we go left two units for our X value and we go down one half of the unit for our Y value of negative one half. That's gonna be approximately here. And then for our point of negative 14, from the origin, we go to the left one unit for X value. And then we're going to go up four units for our Y value of four. That's approximately here. We're very zoomed in on the X axis and zoomed out for the Y axis because of this next one, our point negative 1/4 negative 39.875, we go to the left one half, 1/4 of a unit for our X value and down 39.875 units for our Y value. So not to scale but approximately there. And for a 0.12 from the origin, we go to the right one unit and up two units for our point approximately there. OK. So that's the first six steps. Step seven is just to graph it. So now we can draw the lines connecting all our dots. So we are going to come, we're, we're going to essentially bump up along most of our asom totes. So our first interval from negative infinity going up to our ASOM tote at negative one half. We're going to come along our slant asom tote of Y equals one half, X minus 1/4 from negative infinity for X value. We're going to hit our first point of negative two, negative one half. Then we're going to hit our next point of negative 1.710, our X intercept then curving up hitting our next point of negative 14 and then coming up toward positive infinity for the Y values along our asom tote of X equals negative one half. There's our first one. All right. The second one is going to be between the interval for our Asom toads of negative one half for the X value and X equals zero. And so it has to be below our as our slant asom tote. So this is just gonna hang out down in the third quarter. Our third uh quadrant, excuse me. And it's going to come along our asom tote of X equals negative one half. It's gonna head down that toward negative infinity for the Y value, but we can draw it pops up here and it's going to hit our point at negative 1/4 negative 39.875. And at some point close to that, it's going to turn back around and come down toward negative infinity for the Y value along our vertical Asymptote X equals zero. There's our second line. All right, the third line to be more specific, you could plot more points. But for the sake of time, what this one's going to do is it's going to come, it's, it's heading up toward positive infinity to the right of our asom tote of X equals zero for our Y values. So positive infinity for the Y values and it's coming down and it's going to intersect our point of 12 and then oops, it's going to come along our slant asom tote, heading toward positive infinity for the X values and just bumping up along our slant asom tote of Y equals one half X minus 1/4. OK. All right. That is what our graph is doing. Now, we've got to see which one's the right one. So since we're toward the answers for C and D, I'm going to look at those first. So for answer choice D starting with the last one, they have a slant asom tote of Y equals a negative one half X minus 1/4 but we have a positive one half X minus 1/4. So that one's not going to work. It's actually a very similar graph but that slant Asom tote heads negative, it has a negative slope. So that one doesn't work. That's the part of that graph that doesn't work for answer choice C they have given a slant asom tote of Y equals negative one half X again and then plus 1/4 again, that one doesn't work because of that incorrect slant. Asom tote. And we look at the graph and very similar to what we have. That one reflects it even more closely, but again, incorrect for the slant as some totes, we can rule that one out. So now let's go see what answer choices A and B have. All right, let's make some room for our graph. Bring this up. No. OK. Checking out answer traces A and B. Now answer choices A and B both have the slant asom tote of Y equals one half, X minus 1/4. Great. So those are both in contention. Now, what can we look at and notice in the graph that's different, they both have the vertical asymptotes correct of X equals negative one half and X equals zero. So all of the asymptotes are correct. But what's different is what's going on with the lines themselves just looking at our first line heading from negative infinity for answer choice A that one comes from negative infinity and then it's heading up toward it's from negative infinity for the X value and then heading toward positive infinity for the Y value along our asom tote of X equals negative one half. Whereas the graph for answer choice B is coming from negative infinity for the X value and then heading toward negative infinity for the Y value along our vertical asom tode of X equals negative one half in our answer. Ours goes toward positive infinity along the Asymptote X equals negative one half which matches with answer choice A. So we can rule out answer choice B and we select answer choice. A well done. We'll catch on the next one.

In Exercises 81–88, a. Find the slant asymptote of the graph of each rational function and b. Follow the seven-step strategy and use the slant asymptote to graph each rational function. f(x)=(x³+1)/(x²+2x)

Key Concepts

Slant (Oblique) Asymptotes

Polynomial Long Division

Graphing Rational Functions Using Asymptotes

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In Exercises 81–88, a. Find the slant asymptote of the graph of each rational function and b. Follow the seven-step strategy and use the slant asymptote to graph each rational function. f(x)=(x3+1)/(x2+2x)

Key Concepts

Slant (Oblique) Asymptotes

Polynomial Long Division

Graphing Rational Functions Using Asymptotes

Watch next

In Exercises 81–88, a. Find the slant asymptote of the graph of each rational function and b. Follow the seven-step strategy and use the slant asymptote to graph each rational function. f(x)=(x³+1)/(x²+2x)