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

Question

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

𝓍⁴ ― 1

y = ------------------

𝓍²

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 F of X, which is equal to X, the quantity of X cubed plus 8 in quantity divided by X2. F of X is equal to 1 minus 16 divided by X cubed, and F of X is equal to 48 divided by X4. Below the problem we're given an empty graph on which to plot our function. So, in order to create a graph, we need to analyze our function F of X. So the first thing we need to do is determine the domain of our function. And if we look at F of X, we see that we have a fraction. And that means that our function is going to be defined everywhere except where the denominator is equal to 0, and that's going to occur when X is equal to 0, since our denominator is just X2. So that means that the domain of our function is going to be the open interval from minus infinity to 0. Union with the open interval from 0 to infinity. So our functions are defined everywhere except at X equal to 0. Next, we need to check for symmetries. So for this, we'll calculate f of minus X. And this is going to be equal to the quantity of minus X in quantity cubed, plus 8, and that whole quantity is divided by the quantity of minus X in quantity squared. And simplifying this expression, this is going to be equal to the quantity of minus X cubed plus 8 in quantity, divided by X squared. And this is not equal to F of X. So, we do not have even symmetry, and this is also not equal to minus F of X. So we do not have odd symmetry. So we don't have either even or odd symmetry here, so we can't really use symmetries to create our graph. The next thing we want to look at are asymptotes. And we already saw that our function is undefined whenever X is equal to 0, and that's due to the denominator of a fraction being equal to 0. So that means we're going to have a vertical asytote at X equal to 0. If we look at a fraction, we see that the degree of the polynomial in the numerator is actually one larger than the degree of the denominator. So that means we will not have any horizontal asymptotes, but we will have an oblique or slanted asymptote. So, to define the equation for that line, we're going to take our denominator of X2, and divide it into our numerator of X cubed plus 8. And we're gonna do this by long division. So, X2 will go into X cub X times. So I'm multiply X by X2, I'll get X cubed, and so I'll subtract that from X cube by 8. And when I do that, I get a remainder of 8. Now, I'm not concerned about the remainder, since what I'm interested in for the equation for our oblique asymptote is the quotient. The quotient here gives us the equation for the oblique asymptote, and so that means that Y is going to be equal to X. And so we'll go ahead and draw these asptotes on our graph. So we'll place a vertical dash line at X equal to 0. This will be our vertical aseptote. And then we will have another dash line that follows the equation. Of Y is equal to X, and that's going to be our oblique asmatode. And so we'll just extend. Those lines to the edges of the graph. So now we need to figure out other properties of function F of X in order to draw it on our graph. So the next thing that we want to look at are our critical points, because we'll use those to determine the intervals over which our function is increasing or decreasing. So for this, we'll need to look at our first derivative F of X. I recall that critical points occur whenever our first derivative is undefined or equal to 0. Now, F of X is undefined whenever the denominator in the second term of 1 minus 16 divided by X cubed is equal to 0, and that's going to occur when X is equal to 0. But X equal to 0 is not within the domain of our function, so it cannot be a critical point. So that means that the only critical points are going to be due to setting our first derivative here equal to 0. So we're going to set 1 minus 16 divided by Xecute equal to 0 and solved for X. So the first step is to add 16 divided by X cubed to both sides. Doing so, we'll get 1 is equal to 16 divided by X cubed. Multiplying both sides by X cubed, we'll get X cubed is equal to 16. And then we just need to take the cube root of both sides. So we'll get X is equal to the cube root of 16. Which is approximately equal to 2.52. So now what we're going to do is break up our domain, using our critical points and figure out the intervals over which our function is increasing or decreasing. And to do this, we're going to actually create a table. In the first row of the table, we'll have the sign of F of X. And in the second row, we will have the behavior of F of Xs, whether it's increasing or decreasing. And so the first interval that we're going to want to look over is going from minus infinity to 0. And over this interval, The quantity of -16 divided by X cubed is going to be positive. Since X is negative, X cubed is going to be negative, and when I multiply that by a negative sign, I'll get an overall positive quantity. And so when I add this to 1, I still get a positive quantity. So that means that F of X is going to be positive over this interval. So we'll live with that with a plus sign in our table. And recall that if the first derivative is positive over an interval, that means that our function F of X is 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 from 0 to the cube root of 16, which is our critical point. And on this interval. The quantity of 16 divided by X cubed is actually going to be greater than 1. And so when I subtract that quantity from one, I'm going to get an overall negative quantity. So that means F of X is going to be negative, so we'll put a minus sign in the first row of our table, and recall that if our first derivative is negative over an interval, that means F of X, which in this case our function, is going to be decreasing over that interval. So we'll label that with a D in the second row over our table. And the last example that we want to look at is going from the cube root of 16 to infinity. And over this interval, the quantity of 16 divided by execute is going to be less than 1. Since X cube is always going to be greater than 16, and so when I subtract that quantity from 1, and we could still could get a positive quantity. So that means our first order is positive over this interval, and that means our function F of X is increasing over this interval. So now that we've identified all the intervals over which our function is increasing or decreasing, we see that we have a local minimum at our critical point of X equal to the cube root of 16. Since our function is decreasing in the interval before that and then increasing afterwards. So, the next quantity that we need to look for are our inflection points, and for this we'll need to look at our second derivative, FX. I recall that. Inflection points occur when our second derivative is equal to 0. However, our second derivative here can never be equal to 0. And so that means we don't have any inflection points. Also, our second derivative here is always positive, since 48 is a constant and it's a positive constant in the numerator, and X to the 4th is always a positive quantity, regardless of the value of X. And so, since our second derivative is always positive, that means that our function is going to be concave up over its entire domain. So that means our function is concave up. On the open interval from minus infinity to 0, union with the open interval from 0 to infinity. And so, the last thing we need to do is actually calculate the values of FX when we have our critical points, and we can also look for any other intercepts. So, we're going to calculate F of the cube root of 16. And this is going to be equal to. The cube root of 16, that quantity is going to be cubed. Plus 8 In quantity and that is divided by The quantity of the cube root of 16 in quantity squared. And when we evaluate this expression, this is going to be approximately equal to. 3.78. We can also look for intercepts, and we're actually not going to have any Y intercepts since X equal to 0 is not in the domain of our function. So we just need to look for X intercepts. And so, we're going to need to set the numerator of our function F of X equal to 0, since that's the only time that our function F of X can equal 0, so we'll have X cubed. Plus 8 is equal to 0, and if we solve this for X, we'll get X is equal to -2. So we'll have an X intercept at the point of -2.0. So now we have all the information that we need to actually create our graph. So the first thing I'm going to do is plot the points for our X intercept, which is going to be at -2.0, as well as our local minimum, which is going to occur at 2.52.3.78. So we'll place that point on the graph as well. Now we call that our function is going to be concave up everywhere, and on the interval going from minus infinity to 0, our function is going to be increasing. So we're going to start near the oblique acetode. Our function is going to increase, passing through the X intercept point that we found at X equal to -2, and it's going to continue increasing approaching. The vertical ascitote. Then on the interval from 0 to the cube root of 16, our function is going to be decreasing, still concave up, so we'll start by coming in from positive infinity and coming down near the vertical asptote approaching our local minimum. Add X equal to the cube root of 16, and then after that, our function is going to be increasing, approaching our oblique asymptote. So, the graph that we have drawn on the screen here is what the, is the answer that the problem was looking for. 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.

𝓍⁴ ― 1
y = ------------------
𝓍²

Key Concepts

Rational Functions

Asymptotes

Polynomial Long Division

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