Find the Vertex f(x)=2x^2-3x+6

$f (x) = 2 x^{2} - 3 x + 6$

Step 1

Rewrite the equation in vertex form.

Complete the square for $2 x^{2} - 3 x + 6$ .

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

$a = 2$

$b = - 3$

$c = 6$

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

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

Simplify the right side.

Multiply $2$ by $2$ .

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

Move the negative in front of the fraction.

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

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

$e = 6 - \frac{{(- 3)}^{2}}{4 \cdot 2}$

Simplify the right side.

Simplify each term.

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

$e = 6 - \frac{9}{4 \cdot 2}$

Multiply $4$ by $2$ .

$e = 6 - \frac{9}{8}$

To write $6$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$e = 6 \cdot \frac{8}{8} - \frac{9}{8}$

Combine $6$ and $\frac{8}{8}$ .

$e = \frac{6 \cdot 8}{8} - \frac{9}{8}$

Combine the numerators over the common denominator.

$e = \frac{6 \cdot 8 - 9}{8}$

Simplify the numerator.

Multiply $6$ by $8$ .

$e = \frac{48 - 9}{8}$

Subtract $9$ from $48$ .

$e = \frac{39}{8}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $2 {(x - \frac{3}{4})}^{2} + \frac{39}{8}$ .

$2 {(x - \frac{3}{4})}^{2} + \frac{39}{8}$

Set $y$ equal to the new right side.

$y = 2 {(x - \frac{3}{4})}^{2} + \frac{39}{8}$

Step 2

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

$a = 2$

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

$k = \frac{39}{8}$

Step 3

Find the vertex $(h, k)$ .

$(\frac{3}{4}, \frac{39}{8})$

Step 4

Find the Vertex f(x)=2x^2-3x+6

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