In Exercises 57–80, follow the seven steps to graph each rational...

Question

In Exercises 57–80, follow the seven steps to graph each rational function. f(x)=(x+2)/(x^2+x−6)

Accepted Answer

Hello everybody, everything. All right. Today, today we're going to be looking at this question that states graph the given rational function following the seven steps where the function is F of X is equal to open parentheses, X plus four, closed parentheses, divided by open parentheses, X squared minus X minus 12 closed parentheses. Now the answers provided all consists of different graphs and different versions of what the graph of our F of X function would look like. First, we'll look at a choice A and B which both are separated into three different parts because of a vertical atom tode at X equals negative three and X equals four. And then we'll look at the graphs of C and D which once again are divided into three different parts with vertical asma codes at X equals negative three and X equals four. Now for this question, I'm just going to scroll down to have a white screen where I can use uh the entire screen for my work because we will be dealing with the seven steps which might take a bit of area and a bit of space to work with. So what are the seven steps. The first step here is what we call determining the symmetry. So, all right, the number one for step one and I'll label it as determined symmetry. And how do we do this? Well, there's a couple of rules that we should keep in mind. So those rules are, if F of negative X is equal to F of X, then there will be Y axis symmetry and I'll write sim as an abbreviation for symmetry. Now another rule is if F of negative X is equal to negative F of X, then there will be origin symmetry. So what we have to do is basically use our case and test the F of negative X and see what happens. So for our case, F of negative X is equal to negative X plus four in the numerator and the denominator would consider to be X squared minus X minus 12. Now, this function is neither equal to F of X or negative F of X. Now, when this happens, there will therefore be neither Y axis or origin symmetry. So when neither of the cases is fulfilled, then that just means we won't have symmetry either in the Y axis or the origin. And that'll be your answer for step one. So now we can move on to step two. Now step two is when we start dealing with the intercepts of our graph, we're trying to find out where the intercepts are. So we'll try out the Y intercept first. Now, for the Y intercept, all we have to do is test zero as our X. So we'll have F instead of F of X, we'll have F of zero is equal to. Now we have an enumerator zero plus four divided by zero minus zero minus 12 in the denominator. And this will be equal to four divided by negative 12 which is equal to negative one third or the 0. comma negative one third. That will be R Y intercept zero, comma negative one third. So that is it for step number two. And now we can move on to step number three where we test our other intercept the X intercept. So for the X intercept, it's kind of the same concept instead of testing zero as our X will test zero as our Y. And for our scenario because we have a fraction, all we have to do is set the numerator equal to zero because zero divided by anything is zero. So our new was X plus four. So I have that X plus four equals zero and therefore X is equal to negative four or negative four comma zero as our X intercept. And that is it for step three. So now we can move on to step four where we start talking about the asymptotes and the possible asymptotes in our graph. So we'll look at first the vertical ascent to and I'll abbreviate that as B A. Now for the vertical ascent to all we have to do is set. Now, the denominator equal to zero because if you have a denominator equal to zero, and you're dividing by zero will be undefined, which leads to a vertical ascent to. So our denominator was X squared minus X minus 12 and we said that equal to zero. So now we just have to factor. So in one, we have open parentheses, I just write an open parenthesis and a closed parentheses, multiplying an open parentheses and a closed parentheses. And we'll send that equal to zero. So now we have to fill in the parentheses, we know both will have an X since we're multiplying to get us X squared, now we have to find out two numbers that were multiplied by each other would uh give us negative 12, which is the last term in our equation. But when added with each other will give us negative X which is the middle term. And those two numbers will be positive three and negative four. So we'll have X plus three, multiplying X minus four. So just to check that before we move on, three, multiplied by negative four is negative 12, which is what we wanted and three minus four is negative one X, which is what we wanted. So now we just have to set each parentheses equal to zero because since they're multiplying as long as one of them equals zero, then our final answer will be zero. So we'll have that X plus three equals zero and X minus four equals zero. Therefore, X can be equal to negative three, N X can also be equal to positive four. So we'll have a vertical as to at X equals negative three and X equals four. And that will be our answer for step four. And now we can move on to step five when we look at our other possible as to the horizontal ay. And once again, I'll abbreviate that with H A this time. So for the horizontal asym, so we just have to compare the degrees of the numerator and the denominator. Now to look at the degree, you just look at the power associated the highest power associated with the X, both in the numerator and the denominator. In our case, we have that the green of numerator is one in the degree of denominator is two. Now, why is this? Well, in the numerator, we have an X by itself raised, not raised to any power. So it's X to the first power in the denominator, we have an X squared as our highest power so that the degree will be two. And we know that, so then we know that the degree of the numerator is less than the degree of the denominator. And therefore, this means that there will be a horizontal ascent to at Y equals zero or the X axis. So when you have the degree of the numerator less than the degree of the denominator, there will be a horizontal asom. So at Y equals zero, now we move on to step number six, which is plotting points. Basically, we just want to test different values to have some points to plot in our graph. So for that, I usually do a T chart where I have X values on the left column and Y values on the right column. And here you can test any value you want. I'm going to be testing some specific values that I have written down as for the XS negative 6.828, negative four, negative 1.172 zero five and 10. Now those Y value, those X values will give me a negative 0.682 or negative 6.828 X an X of negative four. We already know will give us a Y of zero. That was our intercept an X of negative 1.172 will give me a Y of negative 0. and X of zero. We already know will give us a Y of negative one third. That was our Y intercept an X of five will give me a Y of 1.125 and an X of 10 will give me a Y of 0.179. So now I'm ready to graph which is step number seven. So step number seven. And our final step will be to graph. So for that, we just write a Y axis and an ax axis. And now we can plot different things that we know. So on the X axis, I'll draw lines at what could be negative two, negative four, negative six. And I'll draw another dash at negative eight. Now, for the positive side, I'll do the same thing. I'll draw one at two four, six and eat. Now for the Y axis, I'll have them more spaced out and I'll draw a dash at one, two and three. And the same can be said for the negative side of the Y axis, I'll draw a negative one, a negative too. Now, first, I'm going to write my vertical asym tilts. So I know I have one at X equals negative three. So I'll go in between the negative four and the negative two and around halfway, I'll draw a dashed line moving up and down my graph and the same will go for X equals four. That was our other vertical Amato. So once again, I'll draw a dashed line up and down my graph to show that there will be vertical as at those points. No, we can put different points that we know. So my X intercept was at negative four comma zero. And then we have a Y intercept at zero comma negative one third. So we go to an X of zero and we go down around a third of the way and we put a point there. Now we have to keep in mind that there is no symmetry. So we will have on the left side, you see that we have an exert to intercept a negative four comma zero. And we also have a point on negative six comma negative 0.682. So a bit under the x axis and we know that we had said that the X X axis was going to be a horizontal as to. So we have to keep in mind there is no symmetry. So it will be a horizontal as until at other points. So now we can complete the left side. We know we have a vertical ascent to at the X of negative three. So if we connect the two points that we have, we see that our graph as we go left is going to be moving very slowly down below the X axis, but continue to move infinitely leftwards. And as we move right past negative four to the right, we have to curve upwards very quickly because we cannot touch X equals negative three. So our graph will curve upwards and move very slowly to the right but very quickly upwards. Now, for the middle portion, we said that we have a point at zero comma negative one third and we have another point at negative one point negative 1.1 72 comma negative 0.299. So this is below or above actually negative one third. So we draw a point slightly above and then we can connect those two points. And as we move to the right of those two points, we're going to have to move, start curving upwards as we approach X equals four because there's a vertical at there. And as we move to the left of that point, we will also have to move or turn downwards because of the vertical ascent to. So now we have kind of what looks like an un upside down parabola in our middle section. Of course, it's not a perfect parabola because it's not symmetrical, but just think of that shape or think of it as an upside down U kind of. So now we can move on to the right side portion of our graph where the horizontal ascent to will be the X axis here. And we can imagine what this is going to look like. So we add the X of five, we are at 1.125 as are why? And at an X of 10, we are almost at zero. So as you can see our graph is curving downwards from those two points and I'm just going to redo that line just so it's more clear. So at an X of five, we are a bit above one and an X of 10 will say around here, we're almost at zero. So we are curving downwards as we move through that graph. And if we move to the left of that, of the two points that we have, we have to meet, curving upwards to not touch our vertical acid to at X equals four. So that will represent a general shape of what a graph will look like. It'll be divided into three parts with vertical ascent tode at X equals negative three and X equals positive four. The middle section will be kind of an upside down U and each right, the left and the right section are almost mirrors of each other. They're not symmetrical, but they are almost mirrors of each other. With the right section being kind of a curved L and the left section being kind of a curved L as well but backwards. So now we just have to look at our answer choices and compare with what we have. And we see that answer choice D has exactly that it has a match with a vertical Aceto at X equals negative three and X equal positive four. With the middle section being kind of an upside down U and the left and right sections being almost mirrors of each other, but not quite. So I hope that was helpful. And until next time.

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In Exercises 57–80, follow the seven steps to graph each rational function. f(x)=(x+2)/(x^2+x−6)

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