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

Question

Eva Abramson · Accepted Answer

Vertical asymptote: At #x=1,-2#

Horizontal asymptote: At #y=0#

X-intercept: None

Y-intercept: At #y=1#

Holes: None

To find the holes and vertical asymptotes, we first study the denominator. We have #x^2+x-2# in the denominator. Factorizing this will give us the vertical asymptotes. #x^2+x-2# #x^2+2x-x-2# #x(x+2)-1(x+2)# #(x+2)(x-1)# The asymptotes are found when the answer is #0#. So, #(x+2)(x-1)=0# #x=1,-2# There are vertical aymptotes at #x=1,-2#. To find the horizontal asymptotes, we must look at the degree of the numerator (#n#) and the denominator (#m#). If #n>m,# there is no horizontal asymptote If #n=m#, we divide the leading coefficients, If #n

Jameson Allen · Answer

undefined To graph the function f(x) = -2/(x^2 + x - 2), we can start by finding the x and y intercepts, identifying any holes, and determining the vertical and horizontal asymptotes.

To find the x-intercepts, we set f(x) equal to zero and solve for x. In this case, there are no x-intercepts.

To find the y-intercept, we substitute x = 0 into the function. The y-intercept is (0, -1).

To identify any holes, we factor the denominator (x^2 + x - 2) and cancel out any common factors with the numerator (-2). In this case, the denominator can be factored as (x - 1)(x + 2). We find that there is a hole at x = 1.

To determine the vertical asymptotes, we set the denominator equal to zero and solve for x. In this case, the vertical asymptotes occur at x = -2 and x = 1.

To find the horizontal asymptote, we examine the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.

Using this information, we can plot the points, draw the vertical asymptotes at x = -2 and x = 1, and draw the horizontal asymptote at y = 0. We also mark the hole at x = 1.