# What is the vertex form of y= 6x^2 + 11x + 4 ?

Nov 29, 2017

the vertex form of the equation is
$y = 6 {\left(x + 0.916666667\right)}^{2} - 1.041666667$

#### Explanation:

The general form of a quadratic equation is
$y = a {x}^{2} + b x + c$

the vertex form of a quadratic equation is
$y = a {\left(x - h\right)}^{2} + k$

where $\left(h , k\right)$ is the vertex of the line

for a standard quadratic the vertex of the line can be found where the slope of the line is equal to 0
The slope of a quadratic is given by the its first derivative
in this case
$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 x + 11$

the slope is $0$ when $x = - \frac{11}{12} \mathmr{and} - 0.916666667$

The original equation
$y = 6 {x}^{2} + 11 x + 4$

Substitute in what we know
$y = 6 \cdot {\left(- \frac{11}{12}\right)}^{2} + 11 \cdot \left(- \frac{11}{12}\right) + 4 = - 1.041666667$

The vertex is at $\left(- 0.916666667 , - 1.041666667\right)$

Thefore
the vertex form of the equation is
$y = 6 {\left(x + 0.916666667\right)}^{2} - 1.041666667$