Question

Show all work to identify the asymptotes and zero of the function f of x equals 5 x over quantity x squared minus 25

Answer 1

Answer: VA: x = {-5, 5), HA: y = 0, Zeros: x = 0

Step-by-step explanation:

$f(x) = \dfrac{5x}{x^2-25}$

Vertical asymptotes: The denominator cannot equal zero.

x² - 25 = 0  

(x - 5)(x + 5) = 0

x = 5, x = -5

Horizontal Asymptote: Let m represent degree of numerator and n represent the degree of the denominator.

1. If m > n then NO asymptote
2. If m = n, then y = coefficient of m ÷ coefficient of n
3. If m < n, then y = 0

For the given equation, m = 1 and n = 2 so we use option 3 --> y = 0

Zeros: Set the equation equal to zero and solve for x.

$0=\dfrac{5x}{x^2-25}$

Multiply both sides by x²- 25   -->    0 = 5x

Divide both sides by 5       -->          0 = x

Answer 2

Using function concepts, it is found that:

• The zero of the function is $x = 0$.

• The vertical asymptotes are $x = \pm 5$.

• The horizontal asymptote is $y = 0$.

The function is given by:

$f(x) = \frac{5x}{x^2 - 25}$

Applying the subtraction of perfect squares, the function will be given by:

$f(x) = \frac{5x}{(x - 5)(x + 5)}$

The zero is the value of x for which:

$f(x) = 0$

In a fraction, it is when the numerator is zero, thus:

$5x = 0$

$x = \frac{0}{5}$

$x = 0$

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator, thus:

$(x - 5)(x + 5) = 0$

$x = \pm 5$

The horizontal asymptote is:

$y = \lim_{x \rightarrow \infty} f(x)$

Hence:

$y = \lim_{x \rightarrow \infty} \frac{5x}{x^2 - 25} = y = \lim_{x \rightarrow \infty} \frac{5x}{x^2} = y = \lim_{x \rightarrow \infty} \frac{5}{x} = 0$

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