Find the Vertex Form x=20y^2-120y+60

$x = 20 y^{2} - 120 y + 60$

Step 1

Complete the square for $20 y^{2} - 120 y + 60$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 20$

$b = - 120$

$c = 60$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 120}{2 \cdot 20}$

Simplify the right side.

Cancel the common factor of $- 120$ and $2$ .

Factor $2$ out of $- 120$ .

$d = \frac{2 \cdot - 60}{2 \cdot 20}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 20$ .

$d = \frac{2 \cdot - 60}{2 (20)}$

Cancel the common factor.

$d = \frac{2 \cdot - 60}{2 \cdot 20}$

Rewrite the expression.

$d = \frac{- 60}{20}$

Cancel the common factor of $- 60$ and $20$ .

Factor $20$ out of $- 60$ .

$d = \frac{20 \cdot - 3}{20}$

Cancel the common factors.

Factor $20$ out of $20$ .

$d = \frac{20 \cdot - 3}{20 (1)}$

Cancel the common factor.

$d = \frac{20 \cdot - 3}{20 \cdot 1}$

Rewrite the expression.

$d = \frac{- 3}{1}$

Divide $- 3$ by $1$ .

$d = - 3$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 60 - \frac{{(- 120)}^{2}}{4 \cdot 20}$

Simplify the right side.

Simplify each term.

Raise $- 120$ to the power of $2$ .

$e = 60 - \frac{14400}{4 \cdot 20}$

Multiply $4$ by $20$ .

$e = 60 - \frac{14400}{80}$

Divide $14400$ by $80$ .

$e = 60 - 1 \cdot 180$

Multiply $- 1$ by $180$ .

$e = 60 - 180$

Subtract $180$ from $60$ .

$e = - 120$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $20 {(y - 3)}^{2} - 120$ .

$20 {(y - 3)}^{2} - 120$

Step 2

Set $x$ equal to the new right side.

$x = 20 {(y - 3)}^{2} - 120$

Find the Vertex Form x=20y^2-120y+60

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