Question

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f(x)= (x^2 + x - 12)/(x^2 + 6x + 9)

Discuss the behavior of f near any excluded x-values/
a. f(x) --> -∞ as x --> 3^+ and as x--> 4^-, f(x) --> ∞ as x --> 4^+ and as x --> 3^-
b. f(x) --> ∞ as x --> 4^+ and as x --> 4^-
c. f(x) --> -∞ as 4^+ and as x --> 4^-, f(x) --> ∞ as 3^+ and as x --> 3^-
d. f(x) --> -∞ as x --> 3^+ and as x --> 3^-, f(x) --> ∞ as x --> -4^+ and as x --> -4^-
e. f(x) --> -∞ as x --> -3^+ and as x --> -3^-

Answer 1

Answer:

  e. f(x) --> -∞ as x --> -3^+ and as x --> -3^-

Step-by-step explanation:

You want to know the behavior of f(x)= (x^2 + x - 12)/(x^2 + 6x + 9) near any excluded x-values.

Domain

The function can be factored as ...

  $f(x)=\dfrac{x^2+x-12}{x^2+6x+9}=\dfrac{(x+4)(x-3)}{(x+3)^2}$

The excluded values are values of x where the denominator is zero. The only excluded value is x = -3. (eliminates all answer choices except E)

Asymptotic behavior

At either side of x = -3, the sign of the numerator is negative and the sign of the denominator is positive. That makes f(3-) < 0 and f(3+) < 0.

f(x) will never approach +∞, but f(x) approaches -∞ as x nears -3 from either direction.

Answer 2

Answer:

$\textsf{e)} \quad f(x) \rightarrow -\infty\;\;\textsf{as}\;\; x \rightarrow -3^+ \;\;\textsf{and as}\;\;x \rightarrow -3^-$

Step-by-step explanation:

Given function:

$f(x)=\dfrac{x^2+x-12}{x^2+6x+9}$

Factor the numerator:

$\implies x^2+x-12$

$\implies x^2+4x-3x-12$

$\implies x(x+4)-3(x+4)$

$\implies (x-3)(x+4)$

Factor the denominator:

$\implies x^2+6x+9$

$\implies x^2+3x+3x+9$

$\implies x(x+3)+3(x+3)$

$\implies (x+3)(x+3)$

$\implies (x+3)^2$

Therefore, the rational function is:

$f(x)=\dfrac{(x-3)(x+4)}{(x+3)^2}$

As the degree of the numerator is equal to the degree of the denominator, there is a horizontal asymptote at y = 1.

A vertical asymptote occurs at the x-value(s) that make the denominator of a rational function zero.

$\implies (x+3)^2=0$

$\implies x+3=0$

$\implies x=-3$

Therefore, there is a vertical asymptote at x = -3.

As there is a vertical asymptote at x = -3, the excluded x-value is x = -3.

As x approaches x = -3 from both sides, the numerator of the rational function approaches -6 and the denominator approaches a very small positive number.  Therefore, the function approaches a very large negative number.

Therefore, the end behaviour of the function as it approaches the excluded value is:

• $f(x) \rightarrow -\infty\;\;\textsf{as}\;\; x \rightarrow -3^+ \;\;\textsf{and as}\;\;x \rightarrow -3^-$