# What is the vertex of  y=3x^2 + 6x + 1 ?

Nov 17, 2015

$\left(- 1 , - 2\right)$

#### Explanation:

Derivate the function and calculate $y ' \left(0\right)$ to find where the the slope is equal to $0$.

$y = 3 {x}^{2} + 6 x + 1$
$y ' = 2 \cdot 3 {x}^{2 - 1} + 1 \cdot 6 {x}^{1 - 0}$
$y ' = 6 x + 6$

Calculate $y ' \left(0\right)$:
$y ' \left(0\right) = 0$
$6 x + 6 = 0$
$6 x = - 6$
$x = - 1$

Put this $x$ value into the original function to find the y-value.
NOTE: Put it in $y$, not $y '$.

$y = 3 \cdot {\left(- 1\right)}^{2} + 6 \cdot \left(- 1\right) + 1$
$y = 3 \cdot 1 - 6 + 1$
$y = 3 - 6 + 1 = - 2$

The vertex is at $\left(- 1 , - 2\right)$