Question

Can someone help me with some algebra?

What is the oblique asymptote of the function f(x) = the quantity x squared plus 5x plus 6 over the quantity x minus 4?

y = x + 9
y = x − 9
y = x + 1
y = x − 1
thanks :)))

Answer 1

★ Oblique Asymptote ★


`f(x) = \frac{x {}^(2) + 5x + 6}{x - 4} \\ f(x) = (x - 4)(x + 9) + 42 \\`
Hence , oblique Asymptote is obtained simultaneously by the Quotient of the function obtained ,
HENCE , oblique Asymptote is

`y = x + 9`

Answer 2

y=x+9.

Explanation:

the function is
`(x^(2)+5x+6)/(x-4)`. Now, the equation of the oblique asymptote is a function y= mx + b where

m =
`\lim_(x \to \infty) (f(x))/(x)`

and b =
`\lim_(x \to \infty) f(x)-mx`

So,

m =
`\lim_(x \to \infty) ((x^(2)+5x+6)/(x-4))/(x)`

=
`\lim_(x \to \infty) (x^(2)+5x+6)/(x^(2)-4x)= 1.`

and,

b =
`\lim_(x \to \infty)(x^(2)+5x+6)/(x-4)-x`

=
`\lim_(x \to \infty)(x^(2)+5x+6-x^(2)+4x)/(x-4)`

=
`\lim_(x \to \infty)(9x+6)/(x-4)= 9.`

Then, the equation of the oblique asymptote is y= x+9.