Find the Vertex x^2-9x+3

$x^{2} - 9 x + 3$

Step 1

Rewrite the equation in vertex form.

Complete the square for $x^{2} - 9 x + 3$ .

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

$a = 1$

$b = - 9$

$c = 3$

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

$d = \frac{- 9}{2 \cdot 1}$

Simplify the right side.

Multiply $2$ by $1$ .

$d = \frac{- 9}{2}$

Move the negative in front of the fraction.

$d = - \frac{9}{2}$

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

$e = 3 - \frac{{(- 9)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 9$ to the power of $2$ .

$e = 3 - \frac{81}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 3 - \frac{81}{4}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = 3 \cdot \frac{4}{4} - \frac{81}{4}$

Combine $3$ and $\frac{4}{4}$ .

$e = \frac{3 \cdot 4}{4} - \frac{81}{4}$

Combine the numerators over the common denominator.

$e = \frac{3 \cdot 4 - 81}{4}$

Simplify the numerator.

Multiply $3$ by $4$ .

$e = \frac{12 - 81}{4}$

Subtract $81$ from $12$ .

$e = \frac{- 69}{4}$

Move the negative in front of the fraction.

$e = - \frac{69}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - \frac{9}{2})}^{2} - \frac{69}{4}$ .

${(x - \frac{9}{2})}^{2} - \frac{69}{4}$

Set $y$ equal to the new right side.

$y = {(x - \frac{9}{2})}^{2} - \frac{69}{4}$

Step 2

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

$a = 1$

$h = \frac{9}{2}$

$k = - \frac{69}{4}$

Step 3

Find the vertex $(h, k)$ .

$(\frac{9}{2}, - \frac{69}{4})$

Step 4

Find the Vertex x^2-9x+3

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