Answer
x-intercept: (1,0)
y-intercept: (0,1)
Vertical Asymptote: x=-1
Domain: (-∞, -1) U (-1, ∞)
Horizontal Asymptote: y=1
Range: [0, ∞)
See graph below
Work Step by Step
$r(x)=\frac{x^2 - 2x + 1}{(x^2 + 2x + 1)}$
$x^2 -2x + 1 = 0$
$(x-1)^2 = 0$
x = 1
x-intercept: (1,0)
y-intercept is the ratio of the constants, which is 1/1 = 1
Thus, the y-intercept is at (0, 1)
Vertical asymptotes are when the denominator is equal to 0
$x^2 + 2x + 1 = 0$
$(x+1)^2 = 0$
x = -1
So, domain is from (-∞, -1) U (-1, ∞)
Horizontal asymptote is the ratio of the constants of the leading term (with equal degree)
1/1 = 1
Thus, the horizontal asymptote is at y=1
So the range is from [0, ∞) (since there is an x-intercept!)
