Question

let g(x)= (2x^{2+}+2x-24) / (x^{3}-16x)

the holes and vertical asymptote(s) for this function exist at?

Step 1

let g(x)= (2x^{2}+2x-24) / (x^{3}-16x)

Let's factrorise the numerator and denominator.

Numerator = 2x^{2}+2x-24 = 2(x^{2}+x-12) = 2(x+4)(x-3)

Denominator = x^{3}-16x = x(x^{2}-16) = x(x-4)(x+4)

Step 2

A hole exists when the numerator and denominator contain the same factor.

The common factor between numerator and denominator is (x+4). Thus there is only one hole and it is located at x + 4 = 0 or x = -4.

Step 3

To find the vertical asymptote(s), find the values of x which make the denominator equal zero, but for which the numerator is not equal to 0.

The denominator is x(x-4)(x+4). The zeroes of the denominator are x = 0, x - 4 = 0 i.e. x = 4 and x + 4 = 0 i.e. x = -4.

Out of these three points, x = -4 is a hole. It's a point at which numerator is also zero. Hence, this point will not be counted as a vertical asympto...

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