Fine the end behavior and x-value of holeEquation is to the right

Question

asked Apr 22, 2023 126k views

1 Answer

Cheroky · Answer 1 · 2023-04-27T22:41:04+0000

Given the function:

$\: f\mleft(x\mright)=(x+1)/(\left(x-3\right)\left(x^2-1\right))$

The end behaviour of the function f(x) describes the behaviour of the function as x approaches +∞ and -∞.

When x approaches +∞,

$\lim _(x\rightarrow\infty)f(x)=\text{ }\lim _(x\rightarrow\infty)(x+1)/((x-3)(x^2-1))=0$

Thus, as x approaches +∞, the function f(x) approaches zero.

When x approaches -∞,

$\lim _(x\rightarrow-\infty)f(x)=\text{ }\lim _(x\rightarrow-\infty)(x+1)/((x-3)(x^2-1))=0$

Thus, as x approaches -∞, the function f(x) approaches zero.

x-value of the hole:

For a rational function f(x) given as

Provided that p(x) and q(x) have a common factor (x-a), the function f(x) will have a hole at x=a.

Thus, from the function f(x)

by expansion, we have

$f(x)=\frac{x+1}{(x-3)(x^{}-1)(x+1)}$

The expression (x-1) is a common factor of the numerator and the denominator.

Thus,

$\begin{gathered} x-1=0 \\ \Rightarrow x=1 \end{gathered}$

Hence, the x-value of the hole is 1.