# How do you find the vertex of  f(x)= -3x^2-6x-7?

Jun 14, 2015

Complete the square to find:

$f \left(x\right) = - 3 {x}^{2} - 6 x - 7 = - 3 {\left(x + 1\right)}^{2} - 4$

has vertex $\left(- 1 , - 4\right)$

#### Explanation:

$f \left(x\right) = - 3 {x}^{2} - 6 x - 7$

$= - 3 {x}^{2} - 6 x - 3 - 4$

$= - 3 \left({x}^{2} + 2 x + 1\right) - 4$

$= - 3 {\left(x + 1\right)}^{2} - 4$

Now ${\left(x + 1\right)}^{2} \ge 0$ has it's minimum value $0$ when $x = - 1$

If $x = - 1$ then

$f \left(x\right) = {\left(x + 1\right)}^{2} - 4 = {0}^{2} - 4 = 0 - 4 = - 4$

So the vertex is at $\left(- 1 , - 4\right)$

In general you can complete the square of any quadratic in $x$ as follows:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$