Find the Vertex y=-x^2-3x

$y = - x^{2} - 3 x$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- x^{2} - 3 x$ .

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

$a = - 1$

$b = - 3$

$c = 0$

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

$d = \frac{- 3}{2 \cdot - 1}$

Simplify the right side.

Multiply $2$ by $- 1$ .

$d = \frac{- 3}{- 2}$

Dividing two negative values results in a positive value.

$d = \frac{3}{2}$

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

$e = 0 - \frac{{(- 3)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 3$ to the power of $2$ .

$e = 0 - \frac{9}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{9}{- 4}$

Move the negative in front of the fraction.

$e = 0 - - \frac{9}{4}$

Multiply $- - \frac{9}{4}$ .

Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{9}{4})$

Multiply $\frac{9}{4}$ by $1$ .

$e = 0 + \frac{9}{4}$

Add $0$ and $\frac{9}{4}$ .

$e = \frac{9}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + \frac{3}{2})}^{2} + \frac{9}{4}$ .

$- {(x + \frac{3}{2})}^{2} + \frac{9}{4}$

Set $y$ equal to the new right side.

$y = - {(x + \frac{3}{2})}^{2} + \frac{9}{4}$

Step 2

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

$a = - 1$

$h = - \frac{3}{2}$

$k = \frac{9}{4}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{3}{2}, \frac{9}{4})$

Step 4

Find the Vertex y=-x^2-3x

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