# What is the vertex form of y= 9x^2 + 27x + 27 ?

Jan 1, 2016

The solution set is: $S = \left\{- \frac{3}{2} , - \frac{27}{4}\right\}$

#### Explanation:

The general formula for a quadratic function is:
$y = A {x}^{2} + B x + C$

To find the vertex, we apply those formulas:
x_(vertex)=−b/(2a)
y_(vertex)=−△/(4a)

In this case:
${x}_{v e r t e x} = - \left(\frac{27}{18}\right) = - \frac{3}{2}$

${y}_{v e r t e x} = - \frac{{27}^{2} - 4 \cdot 9 \cdot 27}{4 \cdot 9}$ To make it easier, we factor the multiples of 3, like this:
${y}_{v e r t e x} = - \frac{{\left({3}^{3}\right)}^{2} - 4 \cdot {3}^{2} \cdot {3}^{3}}{4 \cdot {3}^{2}}$
${y}_{v e r t e x} = - \frac{{3}^{6} - 4 \cdot {3}^{5}}{4 \cdot {3}^{2}} = \frac{{3}^{4} \cdot \cancel{{3}^{2}} - 4 \cdot {3}^{3} \cdot \cancel{{3}^{2}}}{4 \cdot \cancel{{3}^{2}}}$
${y}_{v e r t e x} = - \frac{81 - 108}{4} = - \frac{27}{4}$

So, the solution set is: $S = \left\{- \frac{3}{2} , - \frac{27}{4}\right\}$