Find the Vertex Form x^2+3/2x

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

Step 1

Write $x^{2} + \frac{3}{2} x$ as an equation.

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

Step 2

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

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

Step 3

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

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

$a = 1$

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

$c = 0$

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

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{3}{2} \cdot \frac{1}{2}$ .

Multiply $\frac{3}{2}$ by $\frac{1}{2}$ .

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

Multiply $2$ by $2$ .

$d = \frac{3}{4}$

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

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

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $\frac{3}{2}$ .

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

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

$e = 0 - \frac{\frac{9}{2^{2}}}{4 \cdot 1}$

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

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

Multiply $4$ by $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

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

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

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

Multiply $4$ by $4$ .

$e = 0 - \frac{9}{16}$

Subtract $\frac{9}{16}$ from $0$ .

$e = - \frac{9}{16}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{3}{4})}^{2} - \frac{9}{16}$ .

${(x + \frac{3}{4})}^{2} - \frac{9}{16}$

Step 4

Set $y$ equal to the new right side.

$y = {(x + \frac{3}{4})}^{2} - \frac{9}{16}$

Find the Vertex Form x^2+3/2x

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