Find the Vertex -3x^2+12x-16

$- 3 x^{2} + 12 x - 16$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 3 x^{2} + 12 x - 16$ .

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

$a = - 3$

$b = 12$

$c = - 16$

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

$d = \frac{12}{2 \cdot - 3}$

Simplify the right side.

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

Factor $2$ out of $12$ .

$d = \frac{2 \cdot 6}{2 \cdot - 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

$d = \frac{2 \cdot 6}{2 (- 3)}$

Cancel the common factor.

$d = \frac{2 \cdot 6}{2 \cdot - 3}$

Rewrite the expression.

$d = \frac{6}{- 3}$

Cancel the common factor of $6$ and $- 3$ .

Factor $3$ out of $6$ .

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

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$d = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$d = - 2$

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

$e = - 16 - \frac{12^{2}}{4 \cdot - 3}$

Simplify the right side.

Simplify each term.

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

$e = - 16 - \frac{144}{4 \cdot - 3}$

Multiply $4$ by $- 3$ .

$e = - 16 - \frac{144}{- 12}$

Divide $144$ by $- 12$ .

$e = - 16 - - 12$

Multiply $- 1$ by $- 12$ .

$e = - 16 + 12$

Add $- 16$ and $12$ .

$e = - 4$

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

$- 3 {(x - 2)}^{2} - 4$

Set $y$ equal to the new right side.

$y = - 3 {(x - 2)}^{2} - 4$

Step 2

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

$a = - 3$

$h = 2$

$k = - 4$

Step 3

Find the vertex $(h, k)$ .

$(2, - 4)$

Step 4

Find the Vertex -3x^2+12x-16

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