How do you graph f(x)=1/(x^2-16) using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Jun 15, 2018

See answer below

Explanation:

Given: f(x) = 1/(x^2 - 16)

The given function is a rational (fraction) function of the form:

f(x) = (N(x))/(D(x)) = (a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0)/(b_mx^m + b_(n-1)x^(n-1) + ... + b_1x + b_0)

First factor both the numerator and denominator. The denominator is the difference of squares: (a^2 - b^2) = (a + b)(a - b)

f(x) = 1/(x^2 - 16) = 1/((x +4)(x - 4))

Holes are found where linear terms can be cancelled from the numerator and denominator. In the given function, there are no holes .

Vertical asymptotes are found when D(x) = 0 (Don't include hole factors):

D(x) = (x +4)(x - 4) = 0; " " **x = -4** , " " **x = 4**

Horizontal asymptotes are related to n " and "m:

n < m, horizontal asymptote is y = 0
n = m, horizontal asymptote is y = (a_n)/(b_n)
n > m, no horizontal asymptote

For the given function n = 0, m = 2 => n < m: horizontal asymptote is y = 0

Find intercepts:

Find x - intercepts: set N(x) = 0

No x-intercepts

Find y - intercepts: set x = 0:

f(0) = 1/(0-16) = -1/16

y-intercept: (0, -1/16)