Find the Vertex x^2+4x-12

$x^{2} + 4 x - 12$

Step 1

Rewrite the equation in vertex form.

Complete the square for $x^{2} + 4 x - 12$ .

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

$a = 1$

$b = 4$

$c = - 12$

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

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

Simplify the right side.

Simplify each term.

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

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

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

Cancel the common factors.

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

$e = - 12 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$e = - 12 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = - 12 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = - 12 - 4$

Subtract $4$ from $- 12$ .

$e = - 16$

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

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

Set $y$ equal to the new right side.

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

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 1$

$h = - 2$

$k = - 16$

Step 3

Find the vertex $(h, k)$ .

$(- 2, - 16)$

Step 4

Find the Vertex x^2+4x-12

Related Questions

Support LCMGCF.com by disabling your adblocker.