Find the Vertex 2x-x^2+15

$2 x - x^{2} + 15$

Step 1

Rewrite the equation in vertex form.

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

$y = - x^{2} + 2 x + 15$

Complete the square for $- x^{2} + 2 x + 15$ .

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

$a = - 1$

$b = 2$

$c = 15$

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

$d = \frac{2}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $2$ .

Cancel the common factor.

$d = \frac{2}{2 \cdot - 1}$

Rewrite the expression.

$d = \frac{1}{- 1}$

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

$e = 15 - \frac{2^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = 15 - \frac{4}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 15 - \frac{4}{- 4}$

Divide $4$ by $- 4$ .

$e = 15 - - 1$

Multiply $- 1$ by $- 1$ .

$e = 15 + 1$

Add $15$ and $1$ .

$e = 16$

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

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

Set $y$ equal to the new right side.

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

Step 2

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

$a = - 1$

$h = 1$

$k = 16$

Step 3

Find the vertex $(h, k)$ .

$(1, 16)$

Step 4

Find the Vertex 2x-x^2+15

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