Answer
x-intercept: (-2,0) (1,0)
y-intercept: NONE
Vertical Asymptote: x=-1, 0
Domain: (-∞, -1) U (-1, 0) U (0, ∞)
Horizontal Asymptote: y=2
Range: (-∞, 2) U [18, ∞)
See graph below
Work Step by Step
$r(x)=\frac{2x^2 + 2x - 4}{(x^2 + x)}$
$2x^2 + 2x - 4 = 0$
$2 (x^2 + x - 2) = 0$
$2 (x+2)(x-1) = 0$
x = -2, 1
x-intercept: (-2,0) (1,0)
No y-intercept as when x = 0, the function is undefined
Vertical asymptotes are when the denominator is equal to 0
$x^2 + x = 0$
$x(x+1) = 0$
x = 0, -1
So, domain is from (-∞, -1) U (-1, 0) U (0, ∞)
Horizontal asymptote is the ratio of the constants of the leading term (with equal degree)
2/1 = 2
Thus, the horizontal asymptote is at y=2
So the range is from (-∞, 2) U [18, ∞)
