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

Question

Genesis Allen · Accepted Answer

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# 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)#

Isaiah Anthony · Answer

undefined To graph the function f(x) = (1 - x^2)/x, we can analyze its properties:

1. Holes: The function has a hole at x = 0 since the denominator becomes zero, resulting in an undefined value.

2. Vertical asymptotes: As x approaches positive or negative infinity, the function approaches positive or negative infinity, respectively. Therefore, there are vertical asymptotes at x = -∞ and x = +∞.

3. Horizontal asymptote: To find the horizontal asymptote, we examine the behavior of the function as x approaches positive or negative infinity. By dividing the highest power of x in the numerator and denominator, we see that the degree of the numerator is 2 and the degree of the denominator is 1. Thus, the horizontal asymptote is y = 0.

4. x-intercepts: To find the x-intercepts, we set f(x) = 0 and solve for x. In this case, (1 - x^2)/x = 0 when x = ±1.

5. y-intercept: To find the y-intercept, we substitute x = 0 into the function. In this case, f(0) = (1 - 0^2)/0 = undefined.

By considering these properties, we can graph the function f(x) = (1 - x^2)/x accordingly.