Find the Vertex Form y^2+x-8y-5=0

$y^{2} + x - 8 y - 5 = 0$

Step 1

Move all terms not containing $x$ to the right side of the equation.

Subtract $y^{2}$ from both sides of the equation.

$x - 8 y - 5 = - y^{2}$

Add $8 y$ to both sides of the equation.

$x - 5 = - y^{2} + 8 y$

Add $5$ to both sides of the equation.

$x = - y^{2} + 8 y + 5$

Step 2

Complete the square for $- y^{2} + 8 y + 5$ .

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

$a = - 1$

$b = 8$

$c = 5$

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

$d = \frac{8}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

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

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$d = - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$d = - 4$

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

$e = 5 - \frac{8^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 5 - \frac{64}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 5 - \frac{64}{- 4}$

Divide $64$ by $- 4$ .

$e = 5 - - 16$

Multiply $- 1$ by $- 16$ .

$e = 5 + 16$

Add $5$ and $16$ .

$e = 21$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y - 4)}^{2} + 21$ .

$- {(y - 4)}^{2} + 21$

Step 3

Set $x$ equal to the new right side.

$x = - {(y - 4)}^{2} + 21$

Find the Vertex Form y^2+x-8y-5=0

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