Question

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

Explanation:

Step1: Rewrite the equation for x

Isolate x from $y^{2}-4x + 4y-4=0$. We get $4x=y^{2}+4y - 4$, then $x=\frac{1}{4}y^{2}+y - 1$.

Step2: Analyze the form of the parabola

The general form of a parabola is $x = ay^{2}+by + c$. Here $a=\frac{1}{4}>0$. When $a>0$ in $x = ay^{2}+by + c$, the parabola opens to the right.

Answer:

D. Right