Question

f(x) = x^2+x+1/6(x-4) g(x) = x^2+x+1/8(x-4)

Which statement comparing the oblique asymptotes of the functions f(x) and g(x) is true?


The oblique asymptote for g(x) is steeper than for f(x).

Both functions have the same oblique asymptote.

The oblique asymptote for f(x) is steeper than for g(x).

Both functions have parallel oblique asymptote

Answer 1

Answer:

C, The oblique asymptote for f(x) is steeper than for g(x).

Step-by-step explanation:

Answer 2

Given function $f(x) = \frac{x^{2}+ x + 1}{6(x-4)}$  and  $g(x) = \frac{x^{2}+ x + 1}{8(x-4)}$ true statement which compare the oblique asymptotes of the functions $f(x)$ and $g(x)$ is

the oblique asymptote for $f(x)$ is steeper than for $g(x)$ .

What is oblique asymptote?

" Oblique asymptote is defines as the slant line of the function as x → ∞ and x →(-∞)."

According to the question,

Given function,

$f(x) = \frac{x^{2}+ x + 1}{6(x-4)}$  

 $g(x) = \frac{x^{2}+ x + 1}{8(x-4)}$

Simplify the given function using long division as shown,  

Quotient of both the function is equals to  $x+5$ which is rational , it has oblique asymptote.

As shown in the graph of the function,

Oblique asymptote when x → ∞ and x →(-∞) then $x+5$ →∞ and -∞.

By observation oblique asymptote of $f(x)$is steeper than for $g(x).$

Hence, Option (C) is the correct answer.

Learn more about oblique asymptote here

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