Find the Vertex 4x-x^2-5

$4 x - x^{2} - 5$

Step 1

Rewrite the equation in vertex form.

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

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

Complete the square for $- x^{2} + 4 x - 5$ .

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

$a = - 1$

$b = 4$

$c = - 5$

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

$d = \frac{4}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

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

$e = - 5 - \frac{4^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

$e = - 5 - \frac{4 \cdot 4}{4 \cdot - 1}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = - 5 - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$e = - 5 - - 4$

Multiply $- 1$ by $- 4$ .

$e = - 5 + 4$

Add $- 5$ and $4$ .

$e = - 1$

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

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

Set $y$ equal to the new right side.

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

Step 2

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

$a = - 1$

$h = 2$

$k = - 1$

Step 3

Find the vertex $(h, k)$ .

$(2, - 1)$

Step 4

Find the Vertex 4x-x^2-5

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