Question

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

Answer 1

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)