Determine the vertical asymptote(s)

Question

x³+5x-7/x²-3x-28

i simplified the denominator and got x = -4 and x = 7

x⁵ + 4x³ - 4x + 6/x³ + 2x² - 5x - 6

do i take the denominator and use the quadratic formula to solve for x?

my question is, do i just solve the denominator for the vertical asymptote(s)? or do i also have to figure out the numerator as well?

i'm unsure how to solve these correctly, would appreciate a little help!

Answer 1

You need to verify that the roots are not also roots of the numerator, in which case you might have holes rather than asymptotes.

In the first problem, we know that -4 is not a root, as 4 does not divide 7. With 7, we can verify by substituting that it is not a root.

Thus, x = -4 and x = 7 will be vertical asymptotes.

With the second fraction, we need to go through similar steps. There is a cubic formula, but few use it. From the sign rule, we know that there is at most one positive and 2 negative roots, and we are guaranteed at least one root.

We can see that 2 is a root and use synthetic division. Alternatively, we could just graph and spot the 0s at whole numbers and verify.

2 | 1 2 -5 -6

2 8 6

1 4 3 0

x^2 + 4x +3 = (x+3)(x+1), so we have additional roots -3 and -1

We can now check -3, -1, and 2 with the numerator

x^5 + 4x^3 - 4x + 6; no easy elimination, as 3, 1, and 2 divide 6.

f(-1) = 5;

f(-3) = -33;

f(2) = 62

None of the roots of the denominator is a root of the numerator, so all are vertical asymptotes.

x = -3 x = -1 and x = 2 are vertical asymptotes.

Answer 2

Hi,

x⁵ + 4x³ - 4x + 6/x³ + 2x² - 5x - 6

for finding vertical asymptote you need to solve: x³ + 2x² - 5x - 6 = 0

how can I solve this equation? you need to find all numbers that constant number such 6 can be divided by them ±1,±2,±3,±6 of course it's enough you find one of them, then by helping long division you make denominator such as multiple factor

x³ + 2x² - 5x - 6 = 0 ------> ( x + 1 ) (x + 3 ) ( x - 2 ) = 0 the roots of this equation will be x=-1,x=-3,and x=2

BE CAREFUL, these solution maybe are not vertical asymptote.

How can you be sure x=-1,x=-3,and x=2 are vertical asymptote?

by replacing x = -1 to the nominator and nominator ≠ 0 you can make a decision x = -1 is vertical asymptote then you choose others such as x = -3 and x = 2

so this function f(x) = x⁵ + 4x³ - 4x + 6/(x³ + 2x² - 5x - 6) has three vertical asymptotes x=-1,x=-3,and x=2

I hope it is useful,

Minoo