Find the Standard Form of the Parabola y=-x^2-5x-6

$y = - x^{2} - 5 x - 6$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- x^{2} - 5 x - 6 = y$

Step 2

Add $6$ to both sides of the equation.

$- x^{2} - 5 x = y + 6$

Step 3

Complete the square.

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

$a = - 1$

$b = - 5$

$c = 0 = y + 6$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y + 6$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 5}{2 \cdot - 1}$

Simplify the right side.

Multiply $2$ by $- 1$ .

$d = \frac{- 5}{- 2}$

Dividing two negative values results in a positive value.

$d = \frac{5}{2}$

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

$e = 0 - \frac{{(- 5)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 5$ to the power of $2$ .

$e = 0 - \frac{25}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{25}{- 4}$

Move the negative in front of the fraction.

$e = 0 - - \frac{25}{4}$

Multiply $- - \frac{25}{4}$ .

Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{25}{4})$

Multiply $\frac{25}{4}$ by $1$ .

$e = 0 + \frac{25}{4}$

Add $0$ and $\frac{25}{4}$ .

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

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + \frac{5}{2})}^{2} + \frac{25}{4}$ .

$- {(x + \frac{5}{2})}^{2} + \frac{25}{4} = y + 6$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{25}{4}$ from both sides of the equation.

$- {(x + \frac{5}{2})}^{2} = y + 6 - \frac{25}{4}$

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

$- {(x + \frac{5}{2})}^{2} = y + 6 \cdot \frac{4}{4} - \frac{25}{4}$

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

$- {(x + \frac{5}{2})}^{2} = y + \frac{6 \cdot 4}{4} - \frac{25}{4}$

Combine the numerators over the common denominator.

$- {(x + \frac{5}{2})}^{2} = y + \frac{6 \cdot 4 - 25}{4}$

Simplify the numerator.

Multiply $6$ by $4$ .

$- {(x + \frac{5}{2})}^{2} = y + \frac{24 - 25}{4}$

Subtract $25$ from $24$ .

$- {(x + \frac{5}{2})}^{2} = y + \frac{- 1}{4}$

Move the negative in front of the fraction.

$- {(x + \frac{5}{2})}^{2} = y - \frac{1}{4}$

Step 5

Divide each term in $- {(x + \frac{5}{2})}^{2} = y - \frac{1}{4}$ by $- 1$ and simplify.

Divide each term in $- {(x + \frac{5}{2})}^{2} = y - \frac{1}{4}$ by $- 1$ .

$\frac{- {(x + \frac{5}{2})}^{2}}{- 1} = \frac{y}{- 1} + \frac{- \frac{1}{4}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x + \frac{5}{2})}^{2}}{1} = \frac{y}{- 1} + \frac{- \frac{1}{4}}{- 1}$

Divide ${(x + \frac{5}{2})}^{2}$ by $1$ .

${(x + \frac{5}{2})}^{2} = \frac{y}{- 1} + \frac{- \frac{1}{4}}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{y}{- 1}$ .

${(x + \frac{5}{2})}^{2} = - 1 \cdot y + \frac{- \frac{1}{4}}{- 1}$

Rewrite $- 1 \cdot y$ as $- y$ .

${(x + \frac{5}{2})}^{2} = - y + \frac{- \frac{1}{4}}{- 1}$

Dividing two negative values results in a positive value.

${(x + \frac{5}{2})}^{2} = - y + \frac{\frac{1}{4}}{1}$

Divide $\frac{1}{4}$ by $1$ .

${(x + \frac{5}{2})}^{2} = - y + \frac{1}{4}$

Step 6

Factor.

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

${(x + \frac{5}{2})}^{2} = - y \cdot \frac{4}{4} + \frac{1}{4}$

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

${(x + \frac{5}{2})}^{2} = \frac{- y \cdot 4}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

${(x + \frac{5}{2})}^{2} = \frac{- y \cdot 4 + 1}{4}$

Multiply $4$ by $- 1$ .

${(x + \frac{5}{2})}^{2} = \frac{- 4 y + 1}{4}$

Reorder terms.

${(x + \frac{5}{2})}^{2} = \frac{1}{4} (- 4 y + 1)$

Find the Standard Form of the Parabola y=-x^2-5x-6

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