Find the Vertex y=-1/2x^2-1/4x

$y = - \frac{1}{2} x^{2} - \frac{1}{4} x$

Step 1

Rewrite the equation in vertex form.

Simplify each term.

Combine $x^{2}$ and $\frac{1}{2}$ .

$y = - \frac{x^{2}}{2} - \frac{1}{4} x$

Combine $x$ and $\frac{1}{4}$ .

$y = - \frac{x^{2}}{2} - \frac{x}{4}$

Complete the square for $- \frac{x^{2}}{2} - \frac{x}{4}$ .

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

$a = - \frac{1}{2}$

$b = - \frac{1}{4}$

$c = 0$

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

$d = \frac{- \frac{1}{4}}{2 (- \frac{1}{2})}$

Simplify the right side.

Dividing two negative values results in a positive value.

$d = \frac{\frac{1}{4}}{2 (\frac{1}{2})}$

Multiply the numerator by the reciprocal of the denominator.

$d = \frac{1}{4} \cdot \frac{1}{2 (\frac{1}{2})}$

Combine $2$ and $\frac{1}{2}$ .

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

Divide $2$ by $2$ .

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

Cancel the common factor of $1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$d = \frac{1}{4}$

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

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

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $- \frac{1}{4}$ .

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

Raise $- 1$ to the power of $2$ .

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

Apply the product rule to $\frac{1}{4}$ .

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

One to any power is one.

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

Raise $4$ to the power of $2$ .

$e = 0 - \frac{1 (\frac{1}{16})}{4 (- \frac{1}{2})}$

Multiply $\frac{1}{16}$ by $1$ .

$e = 0 - \frac{\frac{1}{16}}{4 (- \frac{1}{2})}$

Simplify the denominator.

Multiply $- 1$ by $4$ .

$e = 0 - \frac{\frac{1}{16}}{- 4 (\frac{1}{2})}$

Combine $- 4$ and $\frac{1}{2}$ .

$e = 0 - \frac{\frac{1}{16}}{\frac{- 4}{2}}$

Divide $- 4$ by $2$ .

$e = 0 - \frac{\frac{1}{16}}{- 2}$

Multiply the numerator by the reciprocal of the denominator.

$e = 0 - (\frac{1}{16} \cdot \frac{1}{- 2})$

Move the negative in front of the fraction.

$e = 0 - (\frac{1}{16} (- \frac{1}{2}))$

Multiply $\frac{1}{16} (- \frac{1}{2})$ .

Multiply $\frac{1}{16}$ by $\frac{1}{2}$ .

$e = 0 - - \frac{1}{16 \cdot 2}$

Multiply $16$ by $2$ .

$e = 0 - - \frac{1}{32}$

Multiply $- - \frac{1}{32}$ .

Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{1}{32})$

Multiply $\frac{1}{32}$ by $1$ .

$e = 0 + \frac{1}{32}$

Add $0$ and $\frac{1}{32}$ .

$e = \frac{1}{32}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- \frac{1}{2} {(x + \frac{1}{4})}^{2} + \frac{1}{32}$ .

$- \frac{1}{2} {(x + \frac{1}{4})}^{2} + \frac{1}{32}$

Set $y$ equal to the new right side.

$y = - \frac{1}{2} \cdot {(x + \frac{1}{4})}^{2} + \frac{1}{32}$

Step 2

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

$a = - \frac{1}{2}$

$h = - \frac{1}{4}$

$k = \frac{1}{32}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{1}{4}, \frac{1}{32})$

Step 4

Find the Vertex y=-1/2x^2-1/4x

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