Find the Vertex Form y=x^2+14x-5

$y = x^{2} + 14 x - 5$

Step 1

Complete the square for $x^{2} + 14 x - 5$ .

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

$a = 1$

$b = 14$

$c = - 5$

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

$d = \frac{14}{2 \cdot 1}$

Cancel the common factor of $14$ and $2$ .

Factor $2$ out of $14$ .

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 7}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{7}{1}$

Divide $7$ by $1$ .

$d = 7$

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

$e = - 5 - \frac{14^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $14$ to the power of $2$ .

$e = - 5 - \frac{196}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 5 - \frac{196}{4}$

Divide $196$ by $4$ .

$e = - 5 - 1 \cdot 49$

Multiply $- 1$ by $49$ .

$e = - 5 - 49$

Subtract $49$ from $- 5$ .

$e = - 54$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 7)}^{2} - 54$ .

${(x + 7)}^{2} - 54$

Step 2

Set $y$ equal to the new right side.

$y = {(x + 7)}^{2} - 54$