Graph each rational function. See Examples 5–9. ƒ(x)=[(x-5)(x-...

Question

Graph each rational function. See Examples 5–9. ƒ(x)=[(x-5)(x-2)]/(x^2+9)

Accepted Answer

Hello. Today we are going to be drawing the graph for the given rational function. The function that is given to us is F of X is equal to the quantity X minus 11 multiplied by X minus three, divided by X squared plus 169. Now, in order to find the graph of this rational function, there's a few things we first need to identify. We first need to solve for the vertical asymptotes and horizontal asymptotes. We will then need to find the X and Y intercepts and determine the behaviors of the graph. So let's go ahead and start by solving for the vertical asymptotes. In order to solve for the vertical asymptotes, we want to find the restrictions of the domain of the rational function. Now keep in mind that the only value we don't want in the denominator of a rational function is zero. So what we're going to do is we're going to take the denominator which is X squared plus 169 set it equal to zero and solve for X. In order to solve for X, we will take 169 and subtract it to both sides of the equation. This will give us X squared is equal to negative 169. Next, we will take the square root of both sides of the equation. And to be careful because you are taking the square root of a negative number. So that is going to give us X is equal to 13. I. Now, what this means is that we have an imaginary number for X. And since we have an imaginary number, this tells us that there are no vertical asymptotes for the function. So there are no vertical asymptotes. Since there are no vertical asymptotes, we will move on and solve for the horizontal asymptotes. Now, in order to solve for the horizontal asymptotes, we are going to need to look at the highest exponents of the numerator and denominator. In order to do this, we will first have to foil the numerator together. So in the numerator, we are given X minus 11 multiplied by X minus three. In order to simplify this, we are going to have to foil this quantity together. This will give us X squared minus three, X minus 11 X plus 33 and combining like terms together will give us X squared minus 14 X plus 33. This is going to also have the denominator of X squared plus 169. So this is going to be the expanded form of our numerator. Now, what we're going to do is look at the leading exponents of the numerator and denominator. So in order to find the leading exponents notice that the leading exponent or highest exponent of the numerator is two and the highest exponent of the denominator is two as well. So since these exponents are the same, the horizontal Asymptote is going to be the ratio of the leading coefficients of the numerator and denominator. That ratio is going to be one divided by one which will simplify to just be one. So we have a horizontal Asymptote at Y is equal to one. Now that we have the horizontal Asymptote we are going to solve for the X intercept. The X intercepts occur when the function goes to zero. The only value we can't have in the denominator is zero because it becomes undefined. But if the numerator is zero, it causes the entire function to go to zero. So what we are going to do is we are going to take our numerator which is X minus 11 multiplied by X minus three and set it equal to zero. This will give us the values of X that cause the function to go to zero. And that will be our X intercepts using the zero property. We can set the factors of X equal to zero. So we have X minus 11 equal to zero and X minus three equal to zero. And we will solve both of the equations individually on the left hand side, we will add 11 to both sides of the equation. And that will give us X is equal to 11. And on the right hand side, we will add three to both sides of the equation. And that will give us X is equal to three. This tells us that there are two X intercepts, one is located at 11 comma zero and the other is located at three comma zero. Now that we have the X intercepts we are going to solve for the Y intercepts. The white intercepts occur when X is zero. So we are going to plug in the value of zero into our function that will give us F of zero is equal to zero minus 11, multiplied by zero minus three divided by zero squared plus 169. This will simplify to be negative 11 multiplied by negative three divided by 169. The numerator will multiply to give us positive 33 and we are left with 33 divided by 169. This cannot be reduced any further. So the Y intercept will be located at zero comma 33 divided by 169. Now that we have the Y intercepts, we are going to solve for the behaviors of the graph. In order to solve for the behaviors of the graph, we are going to create a number line on this number line. We are going to label the vertical asymptotes and the X intercepts, we didn't have any vertical asymptotes, but we did have two X intercepts. One was located at positive three and the other was located at positive 11. What we are going to do is we are going to take testing values be to the left of positive three between positive three and positive 11 and to the right of positive 11. And depending on what number we get that will tell us if the function increases or decreases in that region. So what's of value to the left of three that we can plug into the function? Well, we can choose zero and we already know that if we plug in zero to the function, we get positive 33 divided by 169. Since this is a positive number, this tells us that the graph will increase to the left of three. A value that we can choose between three and 11 is four. If we plug in four to our function, the output is going to be negative seven divided by 185. Since this output is negative, that tells us that the graph will decrease between three and 11 and a value that we can test to the right of 11 is 12. If we plug in 12 into our function, the output will be nine divided by 313. Since this is a positive number, the graph will increase to the right of positive 11. Now, what we need to do is we need to choose one of the options that best matches all the information we solved. And just to recap, we solved that we get an imaginary number for the vertical Asymptote indicating that there are no vertical asymptotes of the graph. We have a horizontal Asymptote at Y is equal to one. We have X intercepts located at 11 and three on the X axis. We have Y intercepts located at positive 33 divided by 169 on the Y axis. And we know that the graph is going to be increasing to the left of three and to the right of 11 and decreasing between three and 11. The graph that closely matches this option is going to be C and this tells us that the solution to the problem will be C So I hope this video helps you in understanding how to identify the graph of rational function. And I'll go ahead and see you all in the next video.

Graph each rational function. See Examples 5–9. ƒ(x)=[(x-5)(x-2)]/(x^2+9)

Key Concepts

Rational Functions

Graphing Techniques

Asymptotes

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