Find the Standard Form of the Parabola 4x^2-128y=0

$4 x^{2} - 128 y = 0$

Add $128 y$ to both sides of the equation.

$4 x^{2} = 128 y$

Divide each term in $4 x^{2} = 128 y$ by $4$ and simplify.

Divide each term in $4 x^{2} = 128 y$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{128 y}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{128 y}{4}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{128 y}{4}$

Simplify the right side.

Cancel the common factor of $128$ and $4$ .

Factor $4$ out of $128 y$ .

$x^{2} = \frac{4 (32 y)}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$x^{2} = \frac{4 (32 y)}{4 (1)}$

Cancel the common factor.

$x^{2} = \frac{4 (32 y)}{4 \cdot 1}$

Rewrite the expression.

$x^{2} = \frac{32 y}{1}$

Divide $32 y$ by $1$ .

$x^{2} = 32 y$