Question

Which direction does the graph of the equation shown below open?

Answer 1

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 ($y_v$)is calculated as

$y_v = \frac{-b}{2 \times a}$

where a is the coefficient of the quadratic term and b the coefficient of the linear term. Replacing in the formula:

$y_v = \frac{-1}{2 \times 1/4} = -2$

Replacing this value in the quadratic formula we get the x component of the vertex $x_v$

$x_v = y_v^2/4 + y_v - 1$

$x_v = (-2)^2/4 + -2 - 1$

$x_v = -2$

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).

Answer 2

Option A.

you can rewrite the equation as x = y^2 / 4 + y - 1

If you conserve the traditional x-axis and y-axis, the parabola opens to the right (the symetry axis is parallel to x and the function grows as y grows).