How do you find the vertical, horizontal and slant asymptotes of a(x)=(2x^2-1) / (3x^3-2x+1)?

Question

and also for b(x)= (x^3-x+1)/(2x^4+x^3-X^2-1)

what are the x and y intercepts for each?
plz help me asap. thanks

Answer 1

1) Find all the asymptotes of a(x) = (2x² - 1) / (3x³ - 2x + 1).

First factor the denominator.

a(x) = (2x² - 1) / [(x + 1)(3x² - 3x + 1)]

There is one vertical asymptote where the denominator is zero. It is x = -1.

If you take the limit of a(x) as x→∞ you get a horizontal asymptote.

Limit a(x) as x→∞ is 0. The denominator is of greater degree than the numberator.

y = 0 is the horizontal asymptote.

There are no slant asymptotes.

Answer 2

I'll show you how to do a(x) and leave b(x) up to you.



1. Vertical asymptotes - Since the equation of a vertical line is in the form of x = c for some number c, having a vertical asymptote means there is some value for x that is not allowed.

We know that the 11th commandment says, "Thou shalt not divide by zero." So, when you have a rational function (a function that looks like a fraction), the only values of x that breaks the rules are those that make the denominator zero.

In other words, you can find the vertical asymptotes of a rational function by finding the zeros of the denominator. So, in your first problem, find the zeros for (3x^3-2x+1), and you'll find your vertical asymptotes.

Now, to find the zeros of this function, we can take a look at the POSSIBLE rational zeros. Note that the leading coefficient is 3 and the constant term is 1. Possible rational zeros are:

(Factors of 1)/(Factors of 3) = -1, 1, 1/3, and -1/3

Using synthetic division, you can determine that x = -1 is a zero. That's one vertical asymptote. Now, (3x^3-2x+1) factors out to (x+1)(3x^2-3x+1). Using the quadratic formula on the second factor, you'll find that the zeros for that factor are not real. Looks like there is only one vertical asymptote for a(x)...x = -1



2. Horizontal asymptote - The only time there is a horizontal asymptote is when the highest degree of the numerator (let's call it n) is less than or equal to the highest degree of the denominator (let's call this m).

If n < m, then the horizontal asymptote is y = 0.
If n = m, then the horizontal asymptote is y = leading coefficient of the numerator/leading coefficient of the denominator.

For example: f(x) = (3x^2+x-1)/(4x^2-2x+1) has a horizontal asympotote of y = 3/4.

For your problem, n = 2 and m = 3. Since n < m, a(x) has a horizontal asymptote of y = 0.



3. Slant asymptote - The only time there is a slant asymptote is when the highest degree of the numerator is exactly one more than the highest degree of the denominator. Since this isn't the case for a(x), there is no slant asymptote.



4. x-intecept - Note that the x-intercept is where the graph of the function touches the x-axis. This only happens when a(x) = 0. So the question is, when does a rational function ever equal zero? When the numerator is zero! So, set the numerator to zero and solve for it, and you'll find your x-intercepts.

2x^2 - 1 = 0
2x^2 = 1
x^2 = 1/2
x = -1/2, 1/2



5. y-intercept - Note that the y-intercept is where the graph of the function touches the y-axis. This only happens when x = 0. So, set x = 0 and solve for a(0) and you'll find your y - intercepts.

a(0) = [2(0)^2 - 1]/[3(0)^3 - 2(0) + 1]
a(0) = -1/1
a(0) = -1



Hope this helps. Good luck with b(x)!!!

Answer 3

A. whilst the utmost ability of the numerator and the denominator are an identical (right here they are the two x^3), then there's a horizontal asymptote on the line y = a/b the place a is the coefficient in front of the utmost ability of the numerator (as a effect 4) and b is the coefficient in front of the utmost ability of the denominator (as a effect 3). desire this facilitates.

Answer 4

vertical:
you can do it using limits>> divide each term in both the denominator and divisor by x to the highest power - in your case by x^4. next find the limit of b(x) as x approaches infinite. so all the terms that had powers of less than x^4 will become zero and all of the terms with x^4 will be the number of the term itself. so, in your case the vertical asymptote is 0/2 or just x=0.

horizontal:
you take the function in the divisor and set it equal to 0 since for this value b(x) is undefined. then you solve for the roots of that function.

slant: you perform a synthetic division of the function when the divisor is of lower power than the function of the denominator. then, if there is a remainder, your slant equation will be the quotient of your result. if there is no remainder, you have no slant asymptote.

x-intercept:
set y=0 and solve for x.

y-intercept:
set x=0 and solver for y.

Answer 5

You will have to figure out the HA on your own. I will give you the formula. There is something in your math book that will help you. I know that it goes something like if:
m m>n then y=1
m=n then y=o or something like that. The leading coefficient in the numerator is m and the leading coefficient in the denominator is n.


The VA for both problems is all real numbers. I do not remember how to do the SA or the x and y intercepts for these problems. To find the VAs for both set the denominator equal to 0.