Find the Vertex -16x^2+32x+48

$- 16 x^{2} + 32 x + 48$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 16 x^{2} + 32 x + 48$ .

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

$a = - 16$

$b = 32$

$c = 48$

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

$d = \frac{32}{2 \cdot - 16}$

Simplify the right side.

Cancel the common factor of $32$ and $2$ .

Factor $2$ out of $32$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $16$ out of $16$ .

$d = \frac{16 (1)}{- 16}$

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

$d = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$d = - 1$

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

$e = 48 - \frac{32^{2}}{4 \cdot - 16}$

Simplify the right side.

Simplify each term.

Raise $32$ to the power of $2$ .

$e = 48 - \frac{1024}{4 \cdot - 16}$

Multiply $4$ by $- 16$ .

$e = 48 - \frac{1024}{- 64}$

Divide $1024$ by $- 64$ .

$e = 48 - - 16$

Multiply $- 1$ by $- 16$ .

$e = 48 + 16$

Add $48$ and $16$ .

$e = 64$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 16 {(x - 1)}^{2} + 64$ .

$- 16 {(x - 1)}^{2} + 64$

Set $y$ equal to the new right side.

$y = - 16 {(x - 1)}^{2} + 64$

Step 2

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

$a = - 16$

$h = 1$

$k = 64$

Step 3

Find the vertex $(h, k)$ .

$(1, 64)$

Step 4

Find the Vertex -16x^2+32x+48

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