Answer:
A. Right
Step-by-step explanation:
Given
y^2-4x+4y-4=0
Isolating x
y^2 + 4y - 4 = 4x
(y^2)/4 + y - 1 = x
Which is a parabola in the single variable y. That means the curve may open to the right or to the left.
The y component of the vertex of the parabola ([tex] y_v [/tex])is calculated as
[tex] y_v = \frac{-b}{2 \times a} [/tex]
where a is the coefficient of the quadratic term and b the coefficient of the linear term. Replacing in the formula:
[tex] y_v = \frac{-1}{2 \times 1/4} = -2 [/tex]
Replacing this value in the quadratic formula we get the x component of the vertex [tex] x_v [/tex]
[tex] x_v = y_v^2/4 + y_v - 1 [/tex]
[tex] x_v = (-2)^2/4 + -2 - 1 [/tex]
[tex] x_v = -2[/tex]
Now, we have to get another point in the curve. For example, taking y = 0, we get:
x = (y^2)/4 + y - 1
x = (0^2)/4 + 0 - 1
x = -1
Then, the points (-2, -2) and (-1, 0) belong to the parabola, and in consequence, it opens to the right (see figure attached).
