Find the Vertex Form x=y^2-4y+20

$x = y^{2} - 4 y + 20$

Step 1

Complete the square for $y^{2} - 4 y + 20$ .

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

$a = 1$

$b = - 4$

$c = 20$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 4}{2 \cdot 1}$

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

$d = \frac{2 \cdot - 2}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 2}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 2}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 2}{1}$

Divide $- 2$ by $1$ .

$d = - 2$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 20 - \frac{{(- 4)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

Rewrite $- 4$ as $- 1 (4)$ .

$e = 20 - \frac{{(- 1 (4))}^{2}}{4 \cdot 1}$

Apply the product rule to $- 1 (4)$ .

$e = 20 - \frac{{(- 1)}^{2} \cdot 4^{2}}{4 \cdot 1}$

Raise $- 1$ to the power of $2$ .

$e = 20 - \frac{1 \cdot 4^{2}}{4 \cdot 1}$

Multiply $4^{2}$ by $1$ .

$e = 20 - \frac{4^{2}}{4 \cdot 1}$

Factor $4$ out of $4^{2}$ .

$e = 20 - \frac{4 \cdot 4}{4 \cdot 1}$

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

$e = 20 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

$e = 20 - \frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$e = 20 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = 20 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 20 - 4$

Subtract $4$ from $20$ .

$e = 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - 2)}^{2} + 16$ .

${(y - 2)}^{2} + 16$

Step 2

Set $x$ equal to the new right side.

$x = {(y - 2)}^{2} + 16$

Find the Vertex Form x=y^2-4y+20

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