Answer
x-intercept: None
y-intercept: (0, 13/4)
Vertical asymptote: x = 2
Domain: (-∞, 2) U (2, ∞)
Horizontal asymptote: y=3
Range: (3, ∞)
Graph is below.
Work Step by Step
$r(x)=\frac{3x^2-12x + 13}{x^2-4x +4}$
3x^2 - 12x + 13 = 0
3(x^2 - 4x + 4) = -13 + 12
3(x-2)^2 = -1
No x-intercept
y-intercept is the ratio of the constants, which is 13/4
Thus, the y-intercept is at (0, 13/4)
Vertical asymptotes are when the denominator is equal to 0
$x^2 - 4x + 4 = 0$
x = 2
So, the domain is from (-∞, 2) U (2, ∞)
Horizontal asymptote is the ratio of the constants of the leading term (with equal degree)
3/1 = 3
Thus, the horizontal asymptote is at y=3
So the range is from (3, ∞) (no negative values are possible)