Question

For the following function find the vertical, horizontal, and oblique asymptotes, if any. Please show all of your work. f(x) = 4x/7x^2+6x-16 +25.

Can someone please help find the oblique asymptote?! For the following function find the vertical, horizontal, and oblique asymptotes, if any. Please show all of your work. f(x) = 4x/7x^2+6x-16 +25.

Can someone please help find the oblique asymptote?! @Mathematics

Answer 1

Hi, can you tell me what's confusing you, so we can figure out how to work on this together?

Answer 2

Yes I dont know how to find the oblique asymptote

Answer 3

You got it there

\[6x+9\]

Answer 4

Do you understand why?

Answer 5

To find the oblique asymptote, you must use polynomial long division, and then analyze the function as it approaches infinity.

Answer 6

It seems you used the division already. So you have to see what happens when x approach to infinity, and -infinity

Answer 7

what is your function supposed to be

\[f(x) = \frac{4x}{7x^2}+6x-16 +25\]

\[f(x) = \frac{4x}{7x^2+6x-16 +25}\]

\[f(x) = \frac{4x}{7x^2+6x-16 } +25\]

or non of the above?

Answer 8

*none

Answer 9

Zarkon- the last one is what it is supposed to be

Answer 10

This is what I have so far....now I need the interval notation for it....I suppose that is the oblique?

to find the VA you must make the denominator = 0
7x^2+6x−16

factor to find the zeros
(7x-8)(x+2)=0
VA are where this equation = 0
8/7, -2

to find the HA you have to find the limit as
lim x→∞(4x÷(7x2+6x−16))+25

divide both top and bottom by the highest degree in the denominator which is x^2

simplify the expression now that we are left with 4/x in the numerator and 7+(6/x)-(16/x) in the denominator. Now we know that a constant over a infinitely larger variable will go to the y=0 axis. So this means that we are left with 0/7 +25. the equation of the HA is x=25. Because 0/7=0 and 0+25 =25.

Answer 11

the equation is y=25 not x=25

Answer 12

ok

Answer 13

so what is the oblique asymptote?

Answer 14

hello are you there?!