Graph each rational function. ƒ(x)=(4x-2)/(3x+1)

Question

Accepted Answer

Mhm Hey, everyone in this problem, we're asked to draw the graph for the function given below the function we given F of X is equal to five X minus six, divided by eight X plus seven. We're given four answer choices. And they're all graphs. All four of them show a vertical Asymptote at X equals negative one. They have different horizontal asymptotes in each of them. And the direction that the graph follows when approaching the asymptotes is different for each as well. Now we're gonna rewrite the function we were given. So we have F of X is equal to five X minus six divided by eight X plus seven. Let's start with our intercepts. OK. Those give us a good idea what our function looks like. It's two main points we can pick out on our graph. So for our X intercept, we're gonna set Y equal to zero, we get that zero is equal to five X minus six divided by A X plus seven. Now, in order for the right hand side to be equal to zero, we must have that the numerator is equal to zero. So five X minus six must be equal to zero, which tells us that X is equal to six bits. OK? So we have an X intercept, X equals six fits. And so we have the point 6/5 comma zero. Now we're gonna do the same for A Y intercept. Now, the Y intercept, we're gonna set X equals to zero. So we have F of X is equal to zero minus six divided by zero plus seven, which gives us a Y intercept of negative six sevens. And so we have the 0.0, negative 6/7 on our graph. So we already have two points on our graph. Let's keep looking at the information that this function gives us that we can relate to our graph. Now, let's look at whether we have a horizontal asom to OK, the graphs and the answer choices all have horizontal asymptotes at different locations. So let's figure out where this horizontal asom tode should be. Now, when we look at our function, the numerator has a degree of one and the denominator has a degree of one. OK? Remember that when we say degree of a polynomial, it means the highest exponent. OK. They're both linear, the highest exponent is one. So they both have a degree one. So the degree of the numerator is equal to the degree of the denominator. What that tells us is that our horizontal Asymptote is going to be located at the ratio of the leading coefficients. So our horizontal Asymptote is going to be when Y is equal to the leading coefficient of the numerator divided by the leading coefficient of the denominator. Mhm Now, when we say leading coefficient, we mean the coefficient corresponding with the highest degree term. OK. So our highest exponent term is X and so the leading coefficient in the numerator is five, the coefficient corresponding to the X and in the denominator it's a and so we get that our horizontal axon to is that Y is equal to five eights. So we've got our X and Y intercepts and we've got our horizontal Asymptote. Let's look at our vertical Asymptote. Now, our vertical Asymptote is going to occur where the denominator is equal to zero in our function. The denominator is eight X plus seven. OK. So we set this equal to zero we solve for X and we get that X is equal to negative 7/8. So our vertical Asymptote is going to occur but negative seven eights. OK. So let's start drawing this. We have a lot of information. So I'm gonna draw a really um rough sketch of what we have now, our horizontal as to what we found was at 58. So that's gonna be above the X axis. I'm just gonna draw it as a dotted line in blue and this is a line, Y equals 58. Our vertical ascent tode is that X equals negative 78. So on these graphs of our answer choices, it looks like it's at negative one. It's not rate at negative one. It's fairly close and it's actually at negative seven eights, we draw that on our diagram as well and that is gonna be a vertical line. OK. A vertical as is equal to negative seven eights. Now, we can add the two points we found as well. We have the X and Y intercepts, the Y intercept is at zero, negative seven or sorry, negative six sevens know why intercept is a 580. So we have these two points, we have our asymptotes. Now we need to figure out what direction this function is approaching the asymptotes. So let's think about as X goes to our vertical Asymptote negative seven eights and we're gonna start with from the left. OK? And then we're gonna look at from the right. Mm Now, if we approach this vertical asom to from the left, OK? To the left of X equals negative seven eights, the X values are going to be greater or sorry, less than negative seven eights. OK. X values to the left are less than so five X minus six, the X value is gonna be negative. So we have five multiplied by a negative minus six. That factor is gonna be less than zero. We look at eight X plus seven. OK? We have an X value that's slightly less than negative seven eights. So eight X plus seven is also going to be less than zero. OK. So as we get closer and closer to that vertical Asymptote, the denominator is getting closer and closer to zero. On the left hand side, the numerator and denominator are both negative. And so our function F of X is actually positive. OK. A negative divided by a negative gives us a positive. So we know that as the function approaches X equals negative seven eights from the left, it's going to be going to positive infinity like this. Thank we also know that that function is going to reach the horizontal Asymptote as it goes off to positive and negative infinity, it doesn't cross the asymptotes. And so we get that the function will look something like this to the left of that. As now from the right hand side, five X minus six, OK, we're, we still have an X value that's negative because we're just to the right of X equals negative seven eights. So five X minus six is still going to be negative but eight X plus seven is going to be positive now. OK? Because we have an X value that's slightly larger than negative seven eights. That means that our function F FX is going to be negative because we have a negative divided by a positive. So as we approach the vertical AOM to X equals negative 78 from the right, the function is gonna be going off to negative infinity. OK? And we know that as it goes to X equals positive infinity, we're gonna be reaching that horizontal Asymptote. And so our function going through the two points we have is going to look something like this. All right. So using the X and Y intercepts the horizontal Asymptote, the vertical Asymptote and the behavior to the left and right of our vertical Asymptote, we can sketch the graph of this function and we found that it corresponds with option B. Thanks everyone for watching. I hope this video helped see you in the next one.

Graph each rational function. ƒ(x)=(4x-2)/(3x+1)

Key Concepts

Rational Functions

Asymptotes

Intercepts

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