Which direction does the graph of the equation shown below open?
v^2-4x+4y-4=0
a. right
b. down
c. up
d. left

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 ()is calculated as

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

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

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).

1010: 3030

is it i think

are you missing something?

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