Find the Vertex -16x^2+64x+80

$- 16 x^{2} + 64 x + 80$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 16 x^{2} + 64 x + 80$ .

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

$a = - 16$

$b = 64$

$c = 80$

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

$d = \frac{64}{2 \cdot - 16}$

Simplify the right side.

Cancel the common factor of $64$ and $2$ .

Factor $2$ out of $64$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $32$ and $- 16$ .

Factor $16$ out of $32$ .

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

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

$e = 80 - \frac{64^{2}}{4 \cdot - 16}$

Simplify the right side.

Simplify each term.

Raise $64$ to the power of $2$ .

$e = 80 - \frac{4096}{4 \cdot - 16}$

Multiply $4$ by $- 16$ .

$e = 80 - \frac{4096}{- 64}$

Divide $4096$ by $- 64$ .

$e = 80 - - 64$

Multiply $- 1$ by $- 64$ .

$e = 80 + 64$

Add $80$ and $64$ .

$e = 144$

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

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

Set $y$ equal to the new right side.

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

Step 2

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

$a = - 16$

$h = 2$

$k = 144$

Step 3

Find the vertex $(h, k)$ .

$(2, 144)$

Step 4

Find the Vertex -16x^2+64x+80

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