Find the Vertex y=40x-4x^2

$y = 40 x - 4 x^{2}$

Step 1

Rewrite the equation in vertex form.

Reorder $40 x$ and $- 4 x^{2}$ .

$y = - 4 x^{2} + 40 x$

Complete the square for $- 4 x^{2} + 40 x$ .

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

$a = - 4$

$b = 40$

$c = 0$

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

$d = \frac{40}{2 \cdot - 4}$

Simplify the right side.

Cancel the common factor of $40$ and $2$ .

Factor $2$ out of $40$ .

$d = \frac{2 \cdot 20}{2 \cdot - 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4$ .

$d = \frac{2 \cdot 20}{2 (- 4)}$

Cancel the common factor.

$d = \frac{2 \cdot 20}{2 \cdot - 4}$

Rewrite the expression.

$d = \frac{20}{- 4}$

Cancel the common factor of $20$ and $- 4$ .

Factor $4$ out of $20$ .

$d = \frac{4 (5)}{- 4}$

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$d = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$d = - 5$

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

$e = 0 - \frac{40^{2}}{4 \cdot - 4}$

Simplify the right side.

Simplify each term.

Raise $40$ to the power of $2$ .

$e = 0 - \frac{1600}{4 \cdot - 4}$

Multiply $4$ by $- 4$ .

$e = 0 - \frac{1600}{- 16}$

Divide $1600$ by $- 16$ .

$e = 0 - - 100$

Multiply $- 1$ by $- 100$ .

$e = 0 + 100$

Add $0$ and $100$ .

$e = 100$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 4 {(x - 5)}^{2} + 100$ .

$- 4 {(x - 5)}^{2} + 100$

Set $y$ equal to the new right side.

$y = - 4 {(x - 5)}^{2} + 100$

Step 2

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

$a = - 4$

$h = 5$

$k = 100$

Step 3

Find the vertex $(h, k)$ .

$(5, 100)$

Step 4

Find the Vertex y=40x-4x^2

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