Find the Vertex y=3x^2-5x+6

$y = 3 x^{2} - 5 x + 6$

Step 1

Rewrite the equation in vertex form.

Complete the square for $3 x^{2} - 5 x + 6$ .

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

$a = 3$

$b = - 5$

$c = 6$

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

$d = \frac{- 5}{2 \cdot 3}$

Simplify the right side.

Multiply $2$ by $3$ .

$d = \frac{- 5}{6}$

Move the negative in front of the fraction.

$d = - \frac{5}{6}$

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

$e = 6 - \frac{{(- 5)}^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raise $- 5$ to the power of $2$ .

$e = 6 - \frac{25}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 6 - \frac{25}{12}$

To write $6$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$e = 6 \cdot \frac{12}{12} - \frac{25}{12}$

Combine $6$ and $\frac{12}{12}$ .

$e = \frac{6 \cdot 12}{12} - \frac{25}{12}$

Combine the numerators over the common denominator.

$e = \frac{6 \cdot 12 - 25}{12}$

Simplify the numerator.

Multiply $6$ by $12$ .

$e = \frac{72 - 25}{12}$

Subtract $25$ from $72$ .

$e = \frac{47}{12}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x - \frac{5}{6})}^{2} + \frac{47}{12}$ .

$3 {(x - \frac{5}{6})}^{2} + \frac{47}{12}$

Set $y$ equal to the new right side.

$y = 3 {(x - \frac{5}{6})}^{2} + \frac{47}{12}$

Step 2

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

$a = 3$

$h = \frac{5}{6}$

$k = \frac{47}{12}$

Step 3

Find the vertex $(h, k)$ .

$(\frac{5}{6}, \frac{47}{12})$

Step 4

Find the Vertex y=3x^2-5x+6

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