Mathematics, 16.09.2019 08:00 talia43

Which statement is true about the discontinuities of the function f(x)?
f(x)=(x^2-4)/(x^3-x^2-2x)
a. there is a hole at x = 2.
b. there are asymptotes at x = 0 and x = –1.
c. there are asymptotes at x = 0 and x = –1 and a hole at (2, 2/3)
d. there are holes at x = 0 and x = –1 and an asymptote at x = 2.

Answers: 2

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Answer from: makenziehook8

Here we need to see which statements are true about the discontinuities of the given function.

The correct option is b: "There are asymptotes at x = 0 and x = –1."

So a rational function has discontinuities when its denominator becomes equal to zero, this is because we can't divide by zero.

Here the function is:

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

First, we should factorize both numerator and denominator, so we can see if some of the factors appear in both of them and we can remove it.

For the numerator we have:

x^2 - 4 = x^2 - 2^2 = (x + 2)*(x - 2)

For the denominator we have:

x^3 - x^2 - 2x = x*(x^2 - x - 2)

Now we need to find the factorization of the thing inside the parenthesis.

The roots are given by Bhaskara's formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*1*(-2)} }{2*1} \\\\x = \frac{1 \pm 3 }{2}$

So the roots are:

x = (1 + 3)/2 = 2

x = (1 - 3)/2 = -1

And we can rewrite:

x^3 - x^2 - 2x = x*(x^2 - x - 2) = x*(x + 1)*(x - 2)

Then the function can be rewritten as:

$f(x) = \frac{x^2 - 4}{x^3 - x^2 - 2x} = \frac{(x + 2)*(x - 2)}{x*(x + 1)*(x - 2)} = \frac{x + 2}{x*(x + 1)}$

Then the discontinuities are at the zeros of the denominator, which are at x = 0 and x = -1.

Because the denominator goes to zero at these points, we will have vertical asymptotes, thus the correct option is b

"There are asymptotes at x = 0 and x = –1."

If you want to learn more, you can read:

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Answer from: ayoismeisjjjjuan

In order to find out where your holes and asymptotes are is to factor both the top and the bottom of that rational function. The numerator factors to (x+2)(x-2) and the denominator factors to x(x+1)(x-2). So since there is an (x-2) in both the numerator and denominator, that is called a removable discontinuity which we also know as a hole. The other factors in the denominator, the x and the (x+1) are vertical asymptotes, or values of x that make the rational function undefined (you're NEVER allowed to have a 0 in the denominator of a fraction!). So your correct choice is c. The way you find the y coordinate for the hole is to plug in 2 for x and solve it for y. No biggie.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

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Which statement is true about the discontinuities of the function f(x)?
f(x)=(x^2-4)/(x^3-x^2...