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Numerade Educator



Problem 68 Hard Difficulty

Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote.

$ y = \dfrac{x^3}{(x + 1)^2} $


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Video Transcript

here we have domain cannot be won and it cannot be won. No, a negative one cannot be negative one because the denominator cannot be zero intercept. 00 There's no symmetry. But we do have asking. Tooks, you have an s in tow. Er we have vertical one. Andy, Slant one. Now the vertical is the easier one to find. The denominator said equal to zero X equals negative. One is a vertical for our slant as Anto consult for it by doing polynomial division. All right, I'll spoil this out. So we have X squared close to X plus one. Okay, here we have X squared into X cute X X Times X squared would normally be X cubed, but we're going to change the sign to cancel the first term. Okay, so the first term cancels x times, times two ex sort of the sign minus two x squared. Next time one's is going to be. And then there was with the sine minus X. Okay. And now we have, um, ex weird in tow, minus two x squared. So minus two minus two times. Um X weird. Bring this down minus two times x squared. But switched the sign positive to next spared. And now negative. Two times two x would be minus for X. But we have to search a sign so positive for X. All right. And then positive two. Okay. And received. This cancels out, and the next term would have in mind would be the ex, but, um, X squared has a higher power than X. So we're done in their division. We'd, uh, remainder, obviously. But this is the important part that we need right now. X minus two is our slam doesn't. And now we can look a, um, boy prime. Why? Prime is X squared times X plus three over X plus one cubed. Set that equal to zero. I'm gonna need a bigger line in that and we get X equals zero. An X equals negative three. Facilitate. This is F prime. Well, then get it. Three. Now, let's say that zero. And remember, we have asked in tow X equals negative one. So let's just say it's right there. All right, so it's gonna be an increasing interval year decreasing here and then increasing here in here, making this a maximum value local maximum value f of negative. Three is going to be approximately negative. 6.75. That's it for the first derivative test. And now we can look at Kong cavity with the second derivative test. I need to make that a little smaller. Okay, so the second derivative test, Why double print this six X over explosive? One to the fourth said it equal to zero, and we get X equals zero. Put that on her first derivative, second derivative test zero. And we see that it's Kong cave down here. Concave up here, making this an inflection point. Okay, so that's it for Colin Cavity. I made this really small because this craft is going to take up. Look, some space. All right, let's make some room. All right. What has our original functions? Move that over here. I'll Circle and Ritter are in blue. Are original function to make more room for the Griff. Mostly gonna be down here, so all right, We haven't intercepted 00 next weekend. Put our ascent oats at negative one X's negative one. And we have a slant. Wonder why equals X minus two. So if we see that this is too 12 and This is one, too. Then we have a slant, as in two going through here, or it's. And now we know that we have the minimum negative. Three negative. 6.75. So let's just say this is negative. Three negative. 6.75. And we know it's gonna be like this because it is a local maximum. It has the curvature to it. Difficult to draw. But there's a curve. All right. Okay. And next we have our Now, how does the graph approach or intercept? Well, we know that it's going to be con cave down before it reaches zero. So it's gonna be down, down, down, approaching, but never touching the ass in tow. And then it's going to come to zero, and then it's gonna become keep up the rest of the way and approach, but never touch our ass in tow.