Find the Vertex Form x=5y^2-20y+15

$x = 5 y^{2} - 20 y + 15$

Step 1

Complete the square for $5 y^{2} - 20 y + 15$ .

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

$a = 5$

$b = - 20$

$c = 15$

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

$d = \frac{- 20}{2 \cdot 5}$

Simplify the right side.

Cancel the common factor of $- 20$ and $2$ .

Factor $2$ out of $- 20$ .

$d = \frac{2 \cdot - 10}{2 \cdot 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 5$ .

$d = \frac{2 \cdot - 10}{2 (5)}$

Cancel the common factor.

$d = \frac{2 \cdot - 10}{2 \cdot 5}$

Rewrite the expression.

$d = \frac{- 10}{5}$

Cancel the common factor of $- 10$ and $5$ .

Factor $5$ out of $- 10$ .

$d = \frac{5 \cdot - 2}{5}$

Cancel the common factors.

Factor $5$ out of $5$ .

$d = \frac{5 \cdot - 2}{5 (1)}$

Cancel the common factor.

$d = \frac{5 \cdot - 2}{5 \cdot 1}$

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

$e = 15 - \frac{{(- 20)}^{2}}{4 \cdot 5}$

Simplify the right side.

Simplify each term.

Raise $- 20$ to the power of $2$ .

$e = 15 - \frac{400}{4 \cdot 5}$

Multiply $4$ by $5$ .

$e = 15 - \frac{400}{20}$

Divide $400$ by $20$ .

$e = 15 - 1 \cdot 20$

Multiply $- 1$ by $20$ .

$e = 15 - 20$

Subtract $20$ from $15$ .

$e = - 5$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $5 {(y - 2)}^{2} - 5$ .

$5 {(y - 2)}^{2} - 5$

Step 2

Set $x$ equal to the new right side.

$x = 5 {(y - 2)}^{2} - 5$

Find the Vertex Form x=5y^2-20y+15

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