Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = (x² + 12)/(2x + 1)

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to graph the function F of X, which is equal to the quantity of X2 + 14 in quantity, divided by the quantity of 4 X plus 1 in quantity. Below the problem we're given an integraph on which to draw our function. Now, we need to create a graph for a function F of X, and we need to first determine some properties of our function. The first thing we'll look at is the domain. And our domain here is going to be the values of X over which our function is defined. Now our function will be defined everywhere except when the the dominator is going to be equal to 0. So to find this um location points, we'll need to set 4 X plus 1 equal to 0, and we need to solve this for X. So subtracting one from both sides and then dividing both sides by 4. This is X is going to be equal to -1 divided by 4. So that means that our domain is going to be the open interval from minus infinity to -1 divided by 4. Union with the open interval from -1 divided by 4 to infinity. So our function is defined everywhere except X equal to -1 divided by 4. And since our function is not defined at X equal to -1 divided by 4, due to our fraction here, F of X having a zero in the denominator, that means that we have a vertical asymptote at X equal to -1 divided by 4. Also, if we look at our function F of X, we see that the numerator is a polynomial, that is one order higher, one degree higher than the polynomial in the denominator. So that means we're going to have an oblique asymptote. So to find that equation for that oblique asymptote, we're going to divide our denominator of 4 X plus 1 into our numerator, which is X2 + 14. So here we'll use long division. And so 4 X will go into X2 X divided by 4 times. And so when I multiply X divided by 4 by the quantity of 4 X plus 1, I get X2d. Plus X divided by 4, and I'll subtract this quantity from X2 + 14. And when I do that, I'll get minus X divided by 4 + 14. And so here, I now need to divide quantity of 4 X + 1 into this new quantity. And so that means that 4 X will go into minus X divided by 4, minus 1 divided by 16 times. And so when I multiply minus 1 divided by 16 by the quantity of 4 X plus 1, I get minus X divided by 4 minus 1 divided by 16, and I'll subtract this quantity from Minus X is divided by 4 + 14, and I'll get 14 + 1 divided by 16, which would be the remainder from our division here, which I can ignore, because what I actually need. Is what I found in. The quotient So here, Y is going to be equal to X divided by 4 minus 1 divided by 16. So this could be our equation for our oblique asymptote. So we can go ahead and draw these ascetotes on our graph. So the first one that we're going to draw is going to be X equal to -1 divided by 4. So we'll draw a vertical line. Or that asymptote And for the oblique aseptote, I'll need to determine two points for our line here. And so we're gonna choose X equal to 4. That means that Y is going to be equal to 15 divided by 16, so we'll plot that point on our graph. And the other point that we're gonna look at is X equal to 2, and then when I substituted in X equal to 2 into our equation for Y, we see that Y is equal to 7 divided by 16, so we'll plot that point as well. So now what I need to do is just connect a line. Between those two points and extend it over our entire graph. Those are our two asymptotes drawn on the graph. So the next thing we need to look at are our critical points, but this we need the first derivative, so we'll calculate F of X, and that's going to be equal to the derivative with respect to X of the quantity of X2 + 14 in quantity, divided by the quantity of 4 X plus 1 in quantity. So to take this derivative, I'll need to use our quotient rule. So this is going to be equal to, and here we call your quotient rule, that you will have your denominator, which is going to be the quantity of 4 X plus 1. In quantity multiplied by the derivative of our numerator. So we'll have the derivative with respect to X of the quantity of X2 + 14, which is to X. Then we'll have minus our numerator, which is the quantity of X2 + 14 in quantity multiplied by the derivative of the denominator. So we'll have the derivative with respect to x of the quantity of 4 X plus 1, which is 4. This whole quantity is divided by Our denominator squared, thus can be the quantity of 4 X plus 1 in quantity squared. So now we just need to simplify this expression. So, we're going to multiply everything out in the numerator, so this is going to be equal to 8 X2 + 2X minus 4 X2 minus 56, and that quantity is divided by the quantity of 4 X plus 1 in quantity squared. To combined like terms, this is going to be equal to 4 X 2 plus 2X minus 56, that quantity is divided by the quantity of 4 X plus 1 in quantity squared. So, in order to determine a critical points, we need to determine the X values which our first derivative is either undefined or when it's equal to 0. Now, our first derivative is undefined when our denominator here is equal to 0. So that means that 4 x + 1 has equal to 0, which we already found. That means that X is going to be equal to -1 divided by 4, which is the location of our vertical asymptotes. To find the other critical points, we'll need to set our numerator equal to 0, so we'll set. 4 X squared plus 2X minus 56 equal to 0. And we need to solve this equation for X. So, we'll divide both sides of the equation by 2, just simplify our expression. So we'll have 2 X squared, plus X minus 28 is equal to 0, and that can actually factor this expression, so we'll have the quantity of 2 X. Minus 7 in quantity multiplied by the quantity of X plus 4 in quantity that's will be equal to 0. So now I just need to set each factor equal to 0. So we'll have 2 X minus 7 equal to 0, and solving for X, we'll get X is equal to 7 divided by 2, which is 3.5. For the other factor, we'll have X + 4 is equal to 0, and that um solving that for X, we'll get X is equal to minus 4. So our critical points are at X equal to -1 divided by 4, X is equal to 7 divided by 2, and X is equal to -4. So we're going to use those critical points to determine the intervals over which our function F of X is increasing or decreasing. So for this, we're going to make a table. So in the first row of a table, we'll have FX and we'll just be looking at the sign of that function. And then the second row in our table will have the behavior of F of X. So, the first interval that we're going to look at is going to be going from minus infinity to X equal to -4. So here I'm going to choose a point in this interval and evaluate our function F of X. So we're gonna choose X equal to -minus 5. So, calculating F of minus 5, we get 34 divided by 192, which is a positive quantity, and we call that if our first derivative is positive over an interval, that means our function F of X is increasing over that interval. So we'll label that with an I in our table. The next interval that we need to look at is from -4 to -1 divided by 4. Again, we'll choose a point in this interval, so we'll choose X equal to -1. And when we calculate F of -1, we get -6. Which is a negative value. And since our first derivative here is negative, recall that means that our function F of X is decreasing on this interval. So, we'll label that with a D in our table. The next interval that we need to look at is going from -1 divided by 4 to 7 divided by 2. So again, we'll choose a point in this interval, so we'll choose X equal to 0. And so F of 0 is equal to -56, so this is a negative value, which means that our function F of X is decreasing on this interval. And the final interval that we need to look at is going from 7 X equal to 7 divided by 2 to infinity. Again, we'll choose a point in this interval. So we will choose X equal to 4, and we calculate F4. We get 16 divided by 289, which is a positive quantity. And so that means our function F of X is increasing over this interval. So now we've determined the intervals over which our function is increasing and decreasing. We also see that we have a local maximum at X equal to -4, since our function is changing from increasing to decreasing at that point. And we also see that we have a local minimum at 7 X equal to 75 by 2, since our function is changing from decreasing to increasing at that point. So the last thing we need to determine is the concavity of our function F of X. So for this, we need to uh calculate our second derivative, so we'll need to calculate F of X. That's gonna be equal to the derivative with respect to X. Of the quantity of 4 X squared plus 2X minus 56 in quantity, divided by the quantity of 4 X plus 1 in quantity squared. So again, we're gonna need to apply our quotient rule to take this derivative, that this is going to be equal to. Our denominator, which is the quantity of 4 X plus 1 in quantity squared. Multiplied by the derivative of the numerator, which is going to be the quantity of 8 X + 2. In quantity, then we'll have minus our numerator, which is the quantity of 4 X squad. My uh plus 2 X. Minus 56 in quantity, that gets multiplied by the derivative of our denominator. So when we calculate that, I'll get 2 multiplied by the quantity of 4 X plus 1 in quantity multiplied by 4. And this whole quantity is divided by The quantity of 4 x plus 1 in quantity. To the 4th. So, in the numerator we can factor out a 4 X + 1, which will cancel out one of the factors in the denominator, and multiplying out what's left. This is going to be equal to. 32 X squared. Plus 16 X + 2. Minus 32 X squared. -16 X. Plus 448 and that whole quantity is divided by the quantity of 4 x plus 1 in quantity cubed. Simplifying this expression, this is going to be equal to 450 divided by the quantity of 4 x plus 1 in quantity cubed. So now, we need to determine the intervals over which our function F of X is concave up and concave down. For, for this, we're going to make a table, where the first row is going to be the sign of our second derivative of F of X, and the second row is going to be the behavior of F of X. And so our interval here is going to be divided up into two different intervals. So we'll have the first interval going from minus infinity. 2 X equals to minus 1 divided by 4. That's because our denominator will change sign at that point. So for values of X less than -1 divided by 4, our denominator and our second derivative derivative here is negative, which means that overall our second derivative is negative, since our numerator is a positive constant. So that means that F is negative over this interval, which means that our function F of X is concave down. So recall that if our second derivative is negative, that means our function is concave down. Now looking at the interval going from -1 divided by 4 to infinity. Here, our denominator in our second derivative is going to be positive for values of X greater than -1 divided by 4. So that means our second derivative overall is a positive quantity, and that means that our function F of X is going to be concave up on this interval. So now we have all the information that we need to actually draw our graph. So we're gonna labeled, um, our two points are local minima and maxima on our graph. So we're gonna look at X equal to -4, and we calculate F of -4, it has a value of -2. And the other point that we need to calculate is when X is equal to 3.5, and so F of 3.5 is equal to 1.75, so we'll place that point on the graph as well. So now we just need to use the fact that going from minus infinity to -4, our function is increasing and it's concave down. So we'll have our function increasing until we reach the local maximum, which case is going to begin decreasing and approach our vertical asymptotes. Then it will come in from. From positive affinity, along with the vertical acid to 4 X greater than -1 divided by 4, and it's going to decrease until it reaches our local minimum. At X equal to 3.5, and then our function is going to increase as it approaches our asymptote. So, the graph that we have drawn on the screen here is the answer to this question. So, I hope that this explanation has been useful, and I'll see you in the next video.

Graphing functions Use the guidelines of this section to make a complete graph of f.
f(x) = (x² + 12)/(2x + 1)

Key Concepts

Function Analysis

Limits and Asymptotes

Critical Points and Derivatives

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