Find the Vertex y=3x^2+9x

$y = 3 x^{2} + 9 x$

Rewrite the equation in vertex form.

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

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

$a = 3$

$b = 9$

$c = 0$

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

$d = \frac{9}{2 \cdot 3}$

Cancel the common factor of $9$ and $3$ .

Factor $3$ out of $9$ .

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

Cancel the common factors.

Factor $3$ out of $2 \cdot 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

$e = 0 - \frac{9^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{81}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 0 - \frac{81}{12}$

Cancel the common factor of $81$ and $12$ .

Factor $3$ out of $81$ .

$e = 0 - \frac{3 (27)}{12}$

Cancel the common factors.

Factor $3$ out of $12$ .

$e = 0 - \frac{3 \cdot 27}{3 \cdot 4}$

Cancel the common factor.

$e = 0 - \frac{3 \cdot 27}{3 \cdot 4}$

Rewrite the expression.

$e = 0 - \frac{27}{4}$

Subtract $\frac{27}{4}$ from $0$ .

$e = - \frac{27}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x + \frac{3}{2})}^{2} - \frac{27}{4}$ .

$3 {(x + \frac{3}{2})}^{2} - \frac{27}{4}$

Set $y$ equal to the new right side.

$y = 3 {(x + \frac{3}{2})}^{2} - \frac{27}{4}$

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

$a = 3$

$h = - \frac{3}{2}$

$k = - \frac{27}{4}$

Find the vertex $(h, k)$ .

$(- \frac{3}{2}, - \frac{27}{4})$

Find the Vertex y=3x^2+9x

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