Find the Vertex Form x^2-12x+y+40=0

$x^{2} - 12 x + y + 40 = 0$

Step 1

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

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

$- 12 x + y + 40 = - x^{2}$

Add $12 x$ to both sides of the equation.

$y + 40 = - x^{2} + 12 x$

Subtract $40$ from both sides of the equation.

$y = - x^{2} + 12 x - 40$

Step 2

Complete the square for $- x^{2} + 12 x - 40$ .

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

$a = - 1$

$b = 12$

$c = - 40$

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

$d = \frac{12}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $12$ and $2$ .

Factor $2$ out of $12$ .

$d = \frac{2 \cdot 6}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{6}{- 1}$ .

$d = - 1 \cdot 6$

Multiply $- 1$ by $6$ .

$d = - 6$

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

$e = - 40 - \frac{12^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $12$ to the power of $2$ .

$e = - 40 - \frac{144}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 40 - \frac{144}{- 4}$

Divide $144$ by $- 4$ .

$e = - 40 - - 36$

Multiply $- 1$ by $- 36$ .

$e = - 40 + 36$

Add $- 40$ and $36$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 6)}^{2} - 4$ .

$- {(x - 6)}^{2} - 4$

Step 3

Set $y$ equal to the new right side.

$y = - {(x - 6)}^{2} - 4$

Find the Vertex Form x^2-12x+y+40=0

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