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

1 Answer
Jul 6, 2018

no holes
vertical asymptote at x = 0
x-intercepts at (-1, 0), (1, 0)
y-intercept is undefined
no horizontal asymptote
slant/oblique asymptote at y = -x

Explanation:

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

This is a rational function: (N(x))/(D(x)) = (a_nx^n+...)/(b_mx^m+...)

Factor the numerator:
f(x) = -(x^2 -1)/x = -((x-1)(x+1))/x

Holes: Since there are no terms that cancel, there are no holes.

Find vertical asymptotes: when D(x) = 0 => x = 0

Find x-intercepts by setting f(x) = 0:

0 = (1-x^2)/x

0 * x = (1 - x^2)

1 - x^2 = 0

x^2 = 1

x = +- 1

x-intercepts at (-1, 0), (1, 0)

Find y-intercept by setting x = 0 => y = undefined

Find horizontal asymptote by comparing m & n:

If n < m, horizontal asymptote is y = 0

If n = m, horizontal asymptote is y = a_n/b_m

If n > m there is no horizontal aymptote

In the given equation, n = 2, m = 1 => n > m which means there is no horizontal asymptote.

If n = m+1 there is a slant /oblique asymptote.

You need to use long division to find the slant/oblique asymptote :

" "-x
x |bar(-x^2 + 0x + 1)
" "ul(-x^2)
" "0x + 1

" "color(red)(-x) + 1/x
x |bar(-x^2 + 0x + 1)
" "ul(-x^2)
" "0x + 1 This is the remainder

slant/oblique asymptote is at color(red)(y = -x)