Find the Vertex Form y=6x^2+12x-10

$y = 6 x^{2} + 12 x - 10$

Step 1

Complete the square for $6 x^{2} + 12 x - 10$ .

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

$a = 6$

$b = 12$

$c = - 10$

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

$d = \frac{12}{2 \cdot 6}$

Simplify the right side.

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

Factor $2$ out of $12$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 6$ .

$d = \frac{2 \cdot 6}{2 (6)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{6}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$d = \frac{6}{6}$

Rewrite the expression.

$d = 1$

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

$e = - 10 - \frac{12^{2}}{4 \cdot 6}$

Simplify the right side.

Simplify each term.

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

$e = - 10 - \frac{144}{4 \cdot 6}$

Multiply $4$ by $6$ .

$e = - 10 - \frac{144}{24}$

Divide $144$ by $24$ .

$e = - 10 - 1 \cdot 6$

Multiply $- 1$ by $6$ .

$e = - 10 - 6$

Subtract $6$ from $- 10$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $6 {(x + 1)}^{2} - 16$ .

$6 {(x + 1)}^{2} - 16$

Step 2

Set $y$ equal to the new right side.

$y = 6 {(x + 1)}^{2} - 16$

Find the Vertex Form y=6x^2+12x-10

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