Find the Vertex Form -3x^2+x-6

$- 3 x^{2} + x - 6$

Step 1

Write $- 3 x^{2} + x - 6$ as an equation.

$y = - 3 x^{2} + x - 6$

Step 2

Complete the square for $- 3 x^{2} + x - 6$ .

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

$a = - 3$

$b = 1$

$c = - 6$

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

$d = \frac{1}{2 \cdot - 3}$

Simplify the right side.

Multiply $2$ by $- 3$ .

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

Move the negative in front of the fraction.

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

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

$e = - 6 - \frac{1^{2}}{4 \cdot - 3}$

Simplify the right side.

Simplify each term.

One to any power is one.

$e = - 6 - \frac{1}{4 \cdot - 3}$

Multiply $4$ by $- 3$ .

$e = - 6 - \frac{1}{- 12}$

Move the negative in front of the fraction.

$e = - 6 - - \frac{1}{12}$

Multiply $- - \frac{1}{12}$ .

Multiply $- 1$ by $- 1$ .

$e = - 6 + 1 (\frac{1}{12})$

Multiply $\frac{1}{12}$ by $1$ .

$e = - 6 + \frac{1}{12}$

To write $- 6$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$e = - 6 \cdot \frac{12}{12} + \frac{1}{12}$

Combine $- 6$ and $\frac{12}{12}$ .

$e = \frac{- 6 \cdot 12}{12} + \frac{1}{12}$

Combine the numerators over the common denominator.

$e = \frac{- 6 \cdot 12 + 1}{12}$

Simplify the numerator.

Multiply $- 6$ by $12$ .

$e = \frac{- 72 + 1}{12}$

Add $- 72$ and $1$ .

$e = \frac{- 71}{12}$

Move the negative in front of the fraction.

$e = - \frac{71}{12}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 3 {(x - \frac{1}{6})}^{2} - \frac{71}{12}$ .

$- 3 {(x - \frac{1}{6})}^{2} - \frac{71}{12}$

Step 3

Set $y$ equal to the new right side.

$y = - 3 {(x - \frac{1}{6})}^{2} - \frac{71}{12}$

Find the Vertex Form -3x^2+x-6

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