Find the Vertex Form r(x)=(-x^2-2x+1)+(7-x)

$r (x) = (- x^{2} - 2 x + 1) + (7 - x)$

Step 1

Write $r (x) = (- x^{2} - 2 x + 1) + (7 - x)$ as an equation.

$y = (- x^{2} - 2 x + 1) + (7 - x)$

Step 2

Isolate $y$ to the left side of the equation.

Remove parentheses.

$y = - x^{2} - 2 x + 1 + 7 - x$

Simplify $- x^{2} - 2 x + 1 + 7 - x$ .

Subtract $x$ from $- 2 x$ .

$y = - x^{2} - 3 x + 1 + 7$

Add $1$ and $7$ .

$y = - x^{2} - 3 x + 8$

Step 3

Complete the square for $- x^{2} - 3 x + 8$ .

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

$a = - 1$

$b = - 3$

$c = 8$

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

$d = \frac{- 3}{2 \cdot - 1}$

Simplify the right side.

Multiply $2$ by $- 1$ .

$d = \frac{- 3}{- 2}$

Dividing two negative values results in a positive value.

$d = \frac{3}{2}$

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

$e = 8 - \frac{{(- 3)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 3$ to the power of $2$ .

$e = 8 - \frac{9}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 8 - \frac{9}{- 4}$

Move the negative in front of the fraction.

$e = 8 - - \frac{9}{4}$

Multiply $- - \frac{9}{4}$ .

Multiply $- 1$ by $- 1$ .

$e = 8 + 1 (\frac{9}{4})$

Multiply $\frac{9}{4}$ by $1$ .

$e = 8 + \frac{9}{4}$

To write $8$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = 8 \cdot \frac{4}{4} + \frac{9}{4}$

Combine $8$ and $\frac{4}{4}$ .

$e = \frac{8 \cdot 4}{4} + \frac{9}{4}$

Combine the numerators over the common denominator.

$e = \frac{8 \cdot 4 + 9}{4}$

Simplify the numerator.

Multiply $8$ by $4$ .

$e = \frac{32 + 9}{4}$

Add $32$ and $9$ .

$e = \frac{41}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + \frac{3}{2})}^{2} + \frac{41}{4}$ .

$- {(x + \frac{3}{2})}^{2} + \frac{41}{4}$

Step 4

Set $y$ equal to the new right side.

$y = - {(x + \frac{3}{2})}^{2} + \frac{41}{4}$

Find the Vertex Form r(x)=(-x^2-2x+1)+(7-x)

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