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.

what is the original number? you forgot to add it.