Find the Vertex -16x^2+96x

$- 16 x^{2} + 96 x$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 16 x^{2} + 96 x$ .

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

$a = - 16$

$b = 96$

$c = 0$

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

$d = \frac{96}{2 \cdot - 16}$

Simplify the right side.

Cancel the common factor of $96$ and $2$ .

Factor $2$ out of $96$ .

$d = \frac{2 \cdot 48}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$d = \frac{2 \cdot 48}{2 (- 16)}$

Cancel the common factor.

$d = \frac{2 \cdot 48}{2 \cdot - 16}$

Rewrite the expression.

$d = \frac{48}{- 16}$

Cancel the common factor of $48$ and $- 16$ .

Factor $16$ out of $48$ .

$d = \frac{16 (3)}{- 16}$

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

$e = 0 - \frac{96^{2}}{4 \cdot - 16}$

Simplify the right side.

Simplify each term.

Raise $96$ to the power of $2$ .

$e = 0 - \frac{9216}{4 \cdot - 16}$

Multiply $4$ by $- 16$ .

$e = 0 - \frac{9216}{- 64}$

Divide $9216$ by $- 64$ .

$e = 0 - - 144$

Multiply $- 1$ by $- 144$ .

$e = 0 + 144$

Add $0$ and $144$ .

$e = 144$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 16 {(x - 3)}^{2} + 144$ .

$- 16 {(x - 3)}^{2} + 144$

Set $y$ equal to the new right side.

$y = - 16 {(x - 3)}^{2} + 144$

Step 2

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

$a = - 16$

$h = 3$

$k = 144$

Step 3

Find the vertex $(h, k)$ .

$(3, 144)$

Step 4

Find the Vertex -16x^2+96x

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