Question

Where is the function f(x)=3x^2-6x-45/x^2-5x discontinuous, and what types of discontinuities does it have?

Answer 1

Answer:

Below.

Step-by-step explanation:

f(x)=3x^2-6x-45/x^2-5x

= 3x^2-6x-45 / (x(x - 5))

The denominator is zero when x - 0 and x = 5.

So there will be a vertical asymptote  at x = 0 and a hole at x = 5.

If we do the division  we get f(x)  = 3  remainder 9x - 45

so there will also be a horizontal asymptote where x = 3.

Answer 2

Answer:

Hole at (3, 4.8)

Asymptotes at x = 0 and y = 3

Step-by-step explanation:

$f(x)=\dfrac{3x^2-6x-45}{x^2-5x}$

Factor the numerator:

$\implies 3x^2-6x-45$

$\implies 3(x^2-2x-15)$

$\implies 3(x^2+3x-5x-15)$

$\implies 3(x(x+3)-5(x+3))$

$\implies 3(x-5)(x+3)$

Factor the denominator:

$\implies x^2-5x$

$\implies x(x-5)$

Factored form of function:

$\implies f(x)=\dfrac{3(x-5)(x+3)}{x(x-5)}$

Discontinuity: a point at which the function is not continuous.

Holes

After factoring the rational function, if there is a common factor in both the numerator and denominator, there will be a hole at that point (x-value):

$\implies (x-5)=0 \implies x=5$

Factor out the common factor from the function and input the found value of x into the new function to find the y-value of the hole:

$\implies f(x)=\dfrac{3(x+3)}{x}$

$\implies f(5)=\dfrac{3(5+3)}{5}=4.8$

Therefore, there is a hole at (5, 4.8)

Asymptotes

To find the vertical asymptotes, set the denominator of the new (factored) function to zero and solve:

⇒ Vertical Asymptote at x = 0

As the degrees of the leading terms of the numerator and denominator are equal (both x²), the horizontal asymptote is equal to the ratio of the leading coefficients.

⇒ Horizontal Asymptote at y = 3