Find the Vertex Form y=9x^2+9x-1

$y = 9 x^{2} + 9 x - 1$

Complete the square for $9 x^{2} + 9 x - 1$ .

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

$a = 9$

$b = 9$

$c = - 1$

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

$d = \frac{9}{2 \cdot 9}$

Cancel the common factor of $9$ .

Cancel the common factor.

$d = \frac{9}{2 \cdot 9}$

Rewrite the expression.

$d = \frac{1}{2}$

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

$e = - 1 - \frac{9^{2}}{4 \cdot 9}$

Simplify the right side.

Cancel the common factor of $9^{2}$ and $9$ .

Factor $9$ out of $9^{2}$ .

$e = - 1 - \frac{9 \cdot 9}{4 \cdot 9}$

Cancel the common factors.

Factor $9$ out of $4 \cdot 9$ .

$e = - 1 - \frac{9 \cdot 9}{9 \cdot 4}$

Cancel the common factor.

$e = - 1 - \frac{9 \cdot 9}{9 \cdot 4}$

Rewrite the expression.

$e = - 1 - \frac{9}{4}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = - 1 \cdot \frac{4}{4} - \frac{9}{4}$

Combine $- 1$ and $\frac{4}{4}$ .

$e = \frac{- 1 \cdot 4}{4} - \frac{9}{4}$

Combine the numerators over the common denominator.

$e = \frac{- 1 \cdot 4 - 9}{4}$

Simplify the numerator.

Multiply $- 1$ by $4$ .

$e = \frac{- 4 - 9}{4}$

Subtract $9$ from $- 4$ .

$e = \frac{- 13}{4}$

Move the negative in front of the fraction.

$e = - \frac{13}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $9 {(x + \frac{1}{2})}^{2} - \frac{13}{4}$ .

$9 {(x + \frac{1}{2})}^{2} - \frac{13}{4}$

Set $y$ equal to the new right side.

$y = 9 {(x + \frac{1}{2})}^{2} - \frac{13}{4}$

Find the Vertex Form y=9x^2+9x-1

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