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

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

Rewrite the equation in vertex form.

Simplify $- \frac{3}{2} x^{2}$ .

Combine $x^{2}$ and $\frac{3}{2}$ .

$y = - \frac{x^{2} \cdot 3}{2}$

Move $3$ to the left of $x^{2}$ .

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

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

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

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

$b = 0$

$c = 0$

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

$d = \frac{0}{2 (- \frac{3}{2})}$

Simplify the right side.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 (- \frac{3}{2})}$

Cancel the common factors.

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 (- \frac{3}{2})}$

Rewrite the expression.

$d = \frac{0}{- \frac{3}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$d = 0 (- \frac{2}{3})$

Multiply $0 (- \frac{2}{3})$ .

Multiply $- 1$ by $0$ .

$d = 0 (\frac{2}{3})$

Multiply $0$ by $\frac{2}{3}$ .

$d = 0$

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

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

Simplify the right side.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

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

Simplify the denominator.

Multiply $- 1$ by $4$ .

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

Combine $- 4$ and $\frac{3}{2}$ .

$e = 0 - \frac{0}{\frac{- 4 \cdot 3}{2}}$

Multiply $- 4$ by $3$ .

$e = 0 - \frac{0}{\frac{- 12}{2}}$

Divide $- 12$ by $2$ .

$e = 0 - \frac{0}{- 6}$

Divide $0$ by $- 6$ .

$e = 0 - 0$

Multiply $- 1$ by $0$ .

$e = 0 + 0$

Add $0$ and $0$ .

$e = 0$

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

$- \frac{3}{2} x^{2}$

Set $y$ equal to the new right side.

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

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

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

$h = 0$

$k = 0$

Find the vertex $(h, k)$ .

$(0, 0)$

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

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