Find the Vertex 8x^2-x-2

$8 x^{2} - x - 2$

Step 1

Rewrite the equation in vertex form.

Complete the square for $8 x^{2} - x - 2$ .

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

$a = 8$

$b = - 1$

$c = - 2$

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

$d = \frac{- 1}{2 \cdot 8}$

Simplify the right side.

Multiply $2$ by $8$ .

$d = \frac{- 1}{16}$

Move the negative in front of the fraction.

$d = - \frac{1}{16}$

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

$e = - 2 - \frac{{(- 1)}^{2}}{4 \cdot 8}$

Simplify the right side.

Simplify each term.

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

$e = - 2 - \frac{1}{4 \cdot 8}$

Multiply $4$ by $8$ .

$e = - 2 - \frac{1}{32}$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{32}{32}$ .

$e = - 2 \cdot \frac{32}{32} - \frac{1}{32}$

Combine $- 2$ and $\frac{32}{32}$ .

$e = \frac{- 2 \cdot 32}{32} - \frac{1}{32}$

Combine the numerators over the common denominator.

$e = \frac{- 2 \cdot 32 - 1}{32}$

Simplify the numerator.

Multiply $- 2$ by $32$ .

$e = \frac{- 64 - 1}{32}$

Subtract $1$ from $- 64$ .

$e = \frac{- 65}{32}$

Move the negative in front of the fraction.

$e = - \frac{65}{32}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $8 {(x - \frac{1}{16})}^{2} - \frac{65}{32}$ .

$8 {(x - \frac{1}{16})}^{2} - \frac{65}{32}$

Set $y$ equal to the new right side.

$y = 8 {(x - \frac{1}{16})}^{2} - \frac{65}{32}$

Step 2

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

$a = 8$

$h = \frac{1}{16}$

$k = - \frac{65}{32}$

Step 3

Find the vertex $(h, k)$ .

$(\frac{1}{16}, - \frac{65}{32})$

Step 4

Find the Vertex 8x^2-x-2

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