# How do you find the vertex of f(x)= -x^2-3x-6?

Jan 1, 2016

The solution set(or vertex set) is: $S = \left\{- \frac{3}{2} , - \frac{15}{4}\right\}$

#### 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)=−triangle/(4a)

In this case:
${x}_{v e r t e x} = - \frac{- 3}{2 \cdot \left(- 1\right)} = - \left(\frac{3}{2}\right) = - \frac{3}{2}$ and
${y}_{v e r t e x} = - \frac{{\left(- 3\right)}^{2} - 4 \cdot \left(- 1\right) \cdot \left(- 6\right)}{4 \cdot \left(- 1\right)}$
${y}_{v e r t e x} = - \frac{9 - 24}{-} 4 = - \left(\frac{15}{4}\right) = - \frac{15}{4}$

So, the solution set(or vertex set) is: $S = \left\{- \frac{3}{2} , - \frac{15}{4}\right\}$