Show all work to identify the asymptotes and zero of the function f(x)= 5x/x^2-25. Please answer ASAP!!!

The asymptotes and zero of the function are ±5 and 0 respectively

Given the function below;

The vertical asymptotes occur at the point where the function goes large that is the denominator is zero.

If the denominator is zero, hence;

x^2 - 25 = 0

x^2 = 25

x = ±√25

x = ±5

Hence the asymptotes of the function is ±5

For the zero of the function;

Hence the zero of the function is 0

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Answers:

The term "root" is another way of saying "zero of a function"

Explanation:

To find the roots, we need to solve f(x) = 0 for x

f(x) = 0

5x/(x^2 - 25) = 0

5x = 0*(x^2-25)

5x = 0

x = 0/5

x = 0

Plugging x = 0 into f(x) leads to f(x) = 0.

The root is x = 0 meaning the x intercept is at 0 on the x axis number line. This is the origin (0,0).

To find the vertical asymptotes, set the denominator equal to 0 and solve for x

x^2 - 25 = 0

x^2 - 5^2 = 0

(x-5)(x+5) = 0 ... difference of squares rule

x-5 = 0 or x+5 = 0

x = 5 or x = -5

If either x = 5 or x = -5, then then the denominator x^2-25 is zero

We cannot divide by zero, so these values are excluded from the domain, and this produces the vertical asymptotes.

The horizontal asymptote is y = 0 because the degree of the numerator is 1, while the degree of the denominator is 2. The denominator's degree is larger, which leads to y = 0. The rules for finding the horizontal asymptote are

For those rules above, m and n are the degrees of the numerator and denominator respectively. The a,b refers to the leading coefficients of the numerator and denominator.