Give the equations of any vertical, horizontal, or oblique asympt...

Question

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. See Example 4. ƒ(x)=(4-3x)/(2x+1)

Accepted Answer

Hey, everyone in this problem for the following function, we're asked to write the equations for the vertical, horizontal or oblique asymptotes. The function we're given is F of X is equal to eight minus five X divided by three, X plus two. We're given four answer choices. Option A, a vertical Asymptote at X equals negative two thirds and a horizontal Asymptote at Y equals negative five thirds. Option B A vertical Asymptote at X equals negative two thirds and a horizontal Asymptote at Y equals five thirds. Option C A vertical Asymptote at X equals two thirds and a horizontal Asymptote at Y equals negative five thirds. And option D A vertical ascent tode at Y equals negative two thirds and a horizontal Asymptote at X equals negative five thirds. We're gonna rewrite our function F of X. So F of X is equal to eight minus five X divided by three X plus two. And we're gonna start with our horizontal acid totes first. So in order to determine if we have a horizontal ay to and what it is, we want to compare the degree of the numerator of our function to the degree of the denominator. Now, the degree of the numerator in this case is equal to one. OK. The degree again, these are polynomials. So in the numerator, we have this polynomial, the degree is the highest exponent and our highest exponent is one. It's the X term. So the degree of the numerator is equal to one. If we look at the denominator, it also has a degree one. OK? The highest exponent is one. And so the degree of the denominator is equal to the degree of the numerator which tells us that we do have a horizontal Asymptote and that horizontal Asymptote is going to be located at Y is equal to the ratio of the leading coefficients. OK. So we have the leading coefficient of the numerator divided by the leading coefficient of the denominator. I want you to notice that this equation for the horizontal asom tote is Y equals and it will always be Y equals OK. The horizontal asom tote like a horizontal line, if you imagine a horizontal line on a graph that has an equation Y equals something, not X equals something, right? All right. So the leading coefficient is going to be the coefficient corresponding with the highest exponent term. OK. You have to be a little bit careful when the order is mixed up. Like in the numerator here, it's not the first coefficient, it's the coefficient corresponding to the highest degree term. OK. So X is our highest exponent term. And So our coefficient is going to be negative five for the denominator. Again, X is our highest exponent term. The coefficient is three. And so we get that our horizontal ascent to is that Y is equal to negative five thirds. All right. So that's great. Now, if we have a horizontal Asymptote, that means we don't have an oblique Asymptote. OK. For an oblique Asymptote, we need the degree of the numerator to be one bigger than the degree of the denominator. And so if we satisfy the condition of a horizontal Asymptote, we don't satisfy that condition for the oblique Asymptote. So we have our horizontal Asymptote. We do not have an oblique Asymptote and we can move on to our vertical Asymptote. So for the vertical asym tope, what we wanna do is we want to look where the denominator is equal to zero, we have to be a little bit careful. We wanna simplify first. So we wanna look at the denominator equal to zero after simplify. So we wanna take our equation, we want to factor, we want to cancel any terms that can cancel and then look where the denominator is equal to zero. Now, if we look at our function, we have eight minus five X divided by three X plus two, we can't simplify or factor any further than that. So we're gonna take that denominator and set it equal to zero. So we have three X plus two is equal to zero. That tells us that three X is equal to negative two. Can we move that uh plus two to the other side by subtracting? And then we divide by three, we get that X is equal to negative two thirds. So we found that we had a horizontal Asymptote at Y is equal to negative five thirds and we had a vertical ascent tote at X is equal to negative two thirds. So this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. See Example 4. ƒ(x)=(4-3x)/(2x+1)