Sketch the graphs of the rational functions in Exercises 53–60...

Question

Sketch the graphs of the rational functions in Exercises 53–60.

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Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to draw the graph of the function given below, and we're given the function F of X is equal to X2 divided by the quantity of X2 minus 9 in quantity. We're also given F of X, which is equal to -18 x divided by the quantity of X2 minus 9 in quantity squad and F of X, which is equal to the quantity of 54 X2 plus 162 in quantity, divided by the quantity of X2 minus 9 in quantity cubed. Well, the problem we're given an empty graph on which to plot our function. Now, in order to draw our graph of F of X, we need to actually analyze FFAX, and the first thing we need to do is determine the domain of our function. And if we look at our function F of X, we see that we have a fraction. So that means that our function is going to be defined whenever the denominator is not equal to 0. So define the locations where our denominator is equal to 0, we'll set that denominator, which is X2 minus 9 equal to 0, and solve for x. So adding 9 to both sides of the equation we'll get X2 equal to 9, and taking the square root of both sides, we'll get X is equal to plus or minus 3. So that means that our function is not defined at X equal to -3, and it's also not defined at X equal to 3. So, our domain is going to be all the values of X besides those two points. So our domain is going to be the open interval from minus infinity. 2 minus 3. Union with the open interval from -3 to 3. Union with the open interval from 3 to infinity. Next, we can check for symmetries. And so for that we'll calculate f of minus X to check for even or odd symmetries, and this is going to be equal to the quantity of minus X in quantity squared, and that is divided by the quantity of the quantity of minus X in quantity squad minus 9 in quantity. Simplifying this expression, this is going to be equal to x2 divided by the quantity of X2 minus 9, which is equal to F of X. So since F of minus X is equal to F of X, that means that we have even symmetry. So our function is going to be symmetric about the Y axis. Next, we can look for asymptotes. And we already saw that our function F of X was undefined at X equal to + or minus 3, and that was due to the denominator of our fraction being equal to 0. And so that means we're going to have vertical asptotes at X equal to -3 and X equal to 3. Also, if you look at a function F of X, we see that the degree of our numerator is the same as the degree of our denominator. So that means that we're going to have a horizontal asymptote. So, we need to look at the limit. As x approaches infinity. Of the quantity of X2 divided by the quantity of X2 minus 9 in quantity. And what we're going to do is divide both the numerator and denominator by X2d, so we'll have the limit as X approaches infinity of 1, divided by the quantity of 1 minus 9 divided by X 2 in quantity. And so we take the limit as X approaches infinity, the quantity of 9 divided by X2 will approach 0. So this is going to be equal to 1 divided by 1 minus 0. Which is equal to one. And we can apply the similar reasoning if we look at the limit as x approaches minus infinity of the quantity of X2 divided by the quantity of X2 minus 9 in quantity, and this is also going to be equal to 1 by the same reasoning. So that means we're going to have a horizontal asymptote at Y equal to 1. And so we'll go ahead and draw those asymptotes on our graph. So we'll start off by drawing our horizontal asymptote with a dash line at Y equal to 1. And we'll also draw a vertical asymptote at X equal to -3, with a dash line. And we also have to draw another one at X equal to 3. So we'll have 2 vertical asptodes, as well as 1 horizontal ascetode, and those are all with dashed lines. So now we need to determine the intervals over which our function F of X is increasing or decreasing, and for this we'll need to determine our critical points. And for that we'll need to look at F of X. Now recall that critical points occur whenever our first derivative is equal to zero or undefined. So if we look at FX, which is equal to -18 x divided by the quantity of X2 minus 9 in quantity squad, we see that the denominator is the same denominator that we had for original function F of X except it's squared. So that means it's also going to be undefined when X is equal to + or minus 3, but plus or minus 3 is not contained in the domain of our function F of X. So therefore they cannot be critical points. So therefore, the only critical points are going to occur whenever our first derivative is equal to 0, and that's going to occur whenever the numerator is equal to 0. So we're going to set our numerator, which is -18 X. Equal to 0 and solve for X. So dividing both sides by minus 18, we'll have X is equal to 0. So now what we're going to do is divide up our domain using our critical points and determine whether our function is increasing or decreasing over those intervals. So to do this, we're going to create a table. So our table is going to consist of two rows. The first row is going to be the sign of FX, and the second row is going to be the behavior of F of X, whether it's increasing or decreasing. So, the first interval that we're going to look at is going from minus infinity. 2 minus 3. And over this interval, Our function, our value for X is going to be negative, and the sign of FX is determined entirely by the numerator, which is -18 X. So when X is negative, the quantity of minus 18 X is going to be positive. So that means that F of X is going to have an overall positive sign. So we'll label that with a plus in the first row of our table. And recall that if the first row is positive, that means that our function F of X is going to be increasing over that interval. So we'll label that with an I in the second row of our table. Now, the next interval that we want to look at is going to go from X equal to -3 to X equal to 0. And again, here, our X value is going to be negative. And therefore, the quantity of minus 18 X is going to be positive, and since our first derivative is positive, that means our function F of X is increasing over this interval. The next interval that we want to look at is going from 0 to X equal to 3. Now over this interval, our x value is positive, so the quantity of minus 18 X is going to be negative, which means that F of X is also going to be negative. So we'll lab that with a minus sign in the first row of our table. And recall that if the first derivative is negative, that means that our function F of X is going to be decreasing over that interval. So we'll label that with a D in the second row of our table. And then the final interval that we want to look at is going from X equal to 3 to infinity. And over this interval, X again, here is a positive quantity, so minus 18 X is going to be negative, which means that F of X is negative. And therefore our function F of X is decreasing on this interval. So now we've identified all of the intervals over which our function is increasing or decreasing, we see that we have a local maximum at X equal to 0, since our function is changing from increasing to decreasing at this point. Now we also need to find our inflection points, and for this we'll need our second derivative of X, which is equal to the quantity of 54 X2 plus 162 in quantity divided by the quantity of X2 minus 9 in quantity cubed. And recall that inflection points occur whenever our second derivative is equal to 0. So that's going to occur whenever our numerator is equal to 0. However, our numerator in this case is always going to be a positive quantity. It can never be equal to 0. So therefore we have no inflection points. And so we just need to look for the intervals of concavity that are based upon our domains. We'll be looking over 3 different intervals. And for this, we'll need to look at the sign of F of X. And the sign of F of X is determined entirely by our denominator, since our numerator is always a positive quantity. And in our denominator, we have the quantity of X2 minus sign in quantity cubed. And so whenever the quantity of X2 minus 9 is positive, that means that F is going to be positive. As well as when the quantity of X2 minus 9 is negative, that means that F is also going to be negative. So we're just going to focus on the sign of the quantity of X2 minus 9 to determine the sign of F of X. And for this, we'll again use a table where the first row is going to be the sign of F of X. And the second row is gonna be the behavior of F of X, whether it's concave up or concave down. So, the first interval that we need to look at is going from minus infinity to X equal to -3. Now over this interval. The quantity of X2 is actually going to be greater than 9. And therefore when I subtract 9 from it, I'm actually going to get a positive quantity. And so that means that FXx is going to be positive over this interval, and we call that if our second derivative is positive, that means our function F of X is concave up. So we'll label that with a U in our second row of our table. Now the next interval that we want to look at. Is the interval going from X equal to -3 to 3. And over this interval, the quantity of X2d is going to be less than 9. So X2 minus 9 is actually going to be a negative quantity. So we'll label that with a minus sign in the first row of our table, and recall that if the second derivative is negative over an interval, that means that our function F of X is going to be concave down. So we'll label that with a D. And then the final interval that we want to look at is going to go from X equal to 3 to infinity. And again, over this interval, the quantity of X 2 is going to be greater than 9. And therefore, X2 minus sign is going to be a positive quantity. So we'll label this with a plus sign, and that means that our function F of X is going to be concave up on this interval, so we'll label that with a U. So now we have all the information that we need to actually plot our function. So starting with. The interval going from minus infinity to X equal to -3, we see that our function needs to be concave up and increasing over the interval. So we're going to start coming in from minus infinity, just above our. Horizontal asymptote and we're going to continue, and we're going to change. Going up towards our vertical asptote at X equal to -3. So our function is constantly increasing, starting near our horizontal asymptote and going towards our vertical asymptote. Now for the next section, we're going to be looking at the interval from X equal to -3 to X equal to 3. So we're concave down on this interval, but we have a critical point at X equal to 0, and if we evaluate f of 0, that is going to be equal to 0. So we have our local maximum at the origin. And so we're going to be increasing over the interval from x equal to -3 to 0. So we're going to start coming from minus infinity near our vertical asymptote, at X equal to -3, and our function is going to increase until it reaches our local maximum at X equal to 0, and then our function is going to be decreasing. Approaching our vertical asymptote at X equal to 3, and our function is concave down on that end. Now, for the last interval, we're gonna be looking at X equal to 3 to infinity. And by symmetry, this should mirror the interval going from minus infinity to X equal to -3, since we have even symmetry. And so, we're gonna be coming in from infinity, near the vertical asptode at X equals 3. With our function decreasing, and it's going to continue to decrease approaching the horizontal asymptote at Y equal to 1. And so the graph that we have drawn on the screen here is the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Sketch the graphs of the rational functions in Exercises 53–60.

𝓍²
y = ------------------
𝓍² ― 4

Key Concepts

Rational Functions

Vertical Asymptotes

Horizontal Asymptotes

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