Question

This question is worth 100 points please answer now. Algebra 2. I will mark you brainliest.

Answer 1

Answer:

The vertical asymptotes occur at x = 4 and x = -4

The zero of the function is (0, 0)

Step-by-step explanation:

The given function is $f(x) = \dfrac{4\cdot x}{x^2 - 16}$

Therefore, we have;

$f(x) = \dfrac{4\cdot x}{x^2 - 16} = \dfrac{4\cdot x}{(x - 4) \cdot (x + 4)}$

The asymptote is given at the value of x for which the denominator = 0, therefore, the asymptote are the values of x that make the denominator of the equation equal to zero, which are given as follows

For the asymptote, the denominator = x² - 16 = (x - 4)·(x + 4) = 0

Therefore, the vertical asymptotes are the lines x = 4 and x = -4.

The zero of the function is given as follows;

$f(x) = \dfrac{4\cdot x}{x^2 - 16} = 0$

Therefore, 4·x = (x - 4)·(x + 4) × 0 = 0

x = 0/4 = 0

Therefore, when f(x) = 0, x = 0 and the zero of the function = (0, 0).