Question

Please help me

Given f of x is equal to the quantity x minus 4 end quantity divided by the quantity x squared plus 13x plus 36 end quantity, which of the following is true?

f(x) is decreasing for all x > –9
f(x) is increasing for all x < –9
f(x) is increasing for all x > –4
f(x) is decreasing for all x > –4

Answer 1

• f(x)=x-4/x²+13x+36

One asymptote is

• x-4≠0
• x≠4

If you factor the denominator

rest two asymptotes are -4 and -9

• For x<-9 f(x) is increasing

Hence

• for x>-4 x is decreasing

Option D is correct

Answer 2

Answer:

f(x) is increasing for all x > –4

Step-by-step explanation:

Given function:

$f\:(x)=\dfrac{x-4}{x^2+13x+36}$

Asymptote: a line which the curve gets infinitely close to, but never touches.

As the degree of the denominator is larger than the degree of the numerator, there is a horizontal asymptote at y = 0 for the left part of the curve.

To find the vertical asymptotes, set the denominator to zero and solve for x.

$\implies x^2+13x+36=0$

$\implies x^2+4x+9x+36=0$

$\implies x(x+4)+9(x+4)=0$

$\implies (x+9)(x+4)=0$

$x+4=0 \implies x=-4$

$x+9=0 \implies x=-9$

Therefore, the vertical asymptotes are:

• x = -4
• x = -9

Input x-values either side of the vertical asymptotes into the function:

when x < -9

$f\:(-9.1)=\dfrac{(-9.1)-4}{(-9.1)^2+13(-9.1)+36}=-25.68627...$

when x > -4

$f\:(-3.9)=\dfrac{(-3.9)-4}{(-3.9)^2+13(-3.9)+36}=-15.49019...$

A function is said to be increasing if the y-values increase as the x-values increase.

A function is said to be decreasing if the y-values decrease as the x-values increase.

Since we know that as x < -9 the function approaches zero, and at x = -9.1 the function is -25.7, then the function is decreasing for the domain (-∞, -9).

As the function is zero when x = 4, and f(-3.9)=-15.5, we can conclude that the function is increasing for the domain (-4, ∞)

Therefore, the only true statement is:
f(x) is increasing for all x > –4