Find the Vertex Form f(x)=-2x^2+12x-16

$f (x) = - 2 x^{2} + 12 x - 16$

Step 1

Write $f (x) = - 2 x^{2} + 12 x - 16$ as an equation.

$y = - 2 x^{2} + 12 x - 16$

Step 2

Complete the square for $- 2 x^{2} + 12 x - 16$ .

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

$a = - 2$

$b = 12$

$c = - 16$

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

$d = \frac{12}{2 \cdot - 2}$

Simplify the right side.

Cancel the common factor of $12$ and $2$ .

Factor $2$ out of $12$ .

$d = \frac{2 \cdot 6}{2 \cdot - 2}$

Cancel the common factors.

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

$d = \frac{2 \cdot 6}{2 (- 2)}$

Cancel the common factor.

$d = \frac{2 \cdot 6}{2 \cdot - 2}$

Rewrite the expression.

$d = \frac{6}{- 2}$

Cancel the common factor of $6$ and $- 2$ .

Factor $2$ out of $6$ .

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

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$d = - 3$

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

$e = - 16 - \frac{12^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

Raise $12$ to the power of $2$ .

$e = - 16 - \frac{144}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = - 16 - \frac{144}{- 8}$

Divide $144$ by $- 8$ .

$e = - 16 - - 18$

Multiply $- 1$ by $- 18$ .

$e = - 16 + 18$

Add $- 16$ and $18$ .

$e = 2$

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

$- 2 {(x - 3)}^{2} + 2$

Step 3

Set $y$ equal to the new right side.

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

Find the Vertex Form f(x)=-2x^2+12x-16

Related Questions

Support LCMGCF.com by disabling your adblocker.