Find the Vertex y=x^2+x-6

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

Step 1

Rewrite the equation in vertex form.

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

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

$a = 1$

$b = 1$

$c = - 6$

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

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

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

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

Simplify the right side.

Simplify each term.

One to any power is one.

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

Multiply $4$ by $1$ .

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

To write $- 6$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Combine $- 6$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 6$ by $4$ .

$e = \frac{- 24 - 1}{4}$

Subtract $1$ from $- 24$ .

$e = \frac{- 25}{4}$

Move the negative in front of the fraction.

$e = - \frac{25}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{1}{2})}^{2} - \frac{25}{4}$ .

${(x + \frac{1}{2})}^{2} - \frac{25}{4}$

Set $y$ equal to the new right side.

$y = {(x + \frac{1}{2})}^{2} - \frac{25}{4}$

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 1$

$h = - \frac{1}{2}$

$k = - \frac{25}{4}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{1}{2}, - \frac{25}{4})$

Step 4

Find the Vertex y=x^2+x-6

Related Questions

Support LCMGCF.com by disabling your adblocker.