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

Question

Kennedy Adams · Accepted Answer

graph{(x^2-x-2)/(x^2-2x+1) [-10, 10, -4.64, 5.36]}

holes are whatever factors cancels out from the numerator and denominator. What ever factors are left in the denominator if it can equal 0 whatever x equaled for it to be 0 would be the V.A. There is no H.A because both exponents of the numerator and denominator are the same and their coefficients are 1.

Abigail Allen · Answer

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

1. Holes: To find any holes in the graph, we need to identify values of x that make the numerator and denominator equal to zero. In this case, the numerator factors as (x - 2)(x + 1), and the denominator factors as (x - 1)(x - 1). Therefore, the function has a hole at x = 2.

2. Vertical Asymptotes: Vertical asymptotes occur when the denominator of a rational function equals zero, but the numerator does not. In this case, the denominator (x - 1)(x - 1) equals zero at x = 1. Thus, the function has a vertical asymptote at x = 1.

3. Horizontal Asymptotes: To determine the horizontal asymptote(s), we compare the degrees of the numerator and denominator. In this case, both have a degree of 2. Therefore, we divide the leading coefficients of the numerator and denominator, which gives us 1/1 = 1. Hence, the function has a horizontal asymptote at y = 1.

4. x-intercepts: To find the x-intercepts, we set the numerator equal to zero and solve for x. In this case, (x - 2)(x + 1) = 0, which gives us x = 2 and x = -1 as the x-intercepts.

5. y-intercept: To find the y-intercept, we substitute x = 0 into the function. Thus, f(0) = (-2)/1 = -2, giving us the y-intercept at (0, -2).

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