Find the Vertex Form y=x^2+x-3

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

Step 1

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

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

$a = 1$

$b = 1$

$c = - 3$

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

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

Cancel the common factor of $1$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Simplify each term.

One to any power is one.

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

Multiply $4$ by $1$ .

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

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Combine $- 3$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 3$ by $4$ .

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

Subtract $1$ from $- 12$ .

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

Move the negative in front of the fraction.

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

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

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

Step 2

Set $y$ equal to the new right side.

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