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

Jun 13, 2018

$V = \left(- \frac{3}{2} , - \frac{1}{2}\right)$

#### Explanation:

$V = \left(- \frac{b}{2 a} , - \frac{\Delta}{4 a}\right)$

$\Delta = 36 - 4 \cdot 2 \cdot 4 = 4$

$V = \left(- \frac{6}{4} , - \frac{4}{8}\right)$

Jun 13, 2018

$\left(- \setminus \frac{3}{2} , - \setminus \frac{1}{2}\right)$

#### Explanation:

Method 1: Calculus approach

Vertex is where the gradient of the curve is 0.

Therefore find $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 4 x + 6$

Equate that to 0 such that:

$4 x + 6 = 0$

Solve for $x$, $x = - \setminus \frac{3}{2}$

Let $x = - \setminus \frac{3}{2}$ into the original function, therefore

$y = 2 \cdot {\left(- \setminus \frac{3}{2}\right)}^{2} + 6 \cdot \left(- \setminus \frac{3}{2}\right) + 4$

$y = - \setminus \frac{1}{2}$

Method 2: Algebraic approach.

Complete the square to find the turning points, also known as the vertex.

$y = 2 {x}^{2} + 6 x + 4$
$y = 2 \left({x}^{2} + 3 x + 2\right)$
$y = 2 \left[{\left(x + \setminus \frac{3}{2}\right)}^{2} - \setminus \frac{9}{3} + 2\right]$
$y = 2 {\left(x + \setminus \frac{3}{2}\right)}^{2} - \setminus \frac{1}{2}$

Notice here that you have to multiply BOTH terms by 2, as 2 was the common factor which you took out of the entire expression!

Therefore, the turning points can be picked up such that
$x = - \setminus \frac{3}{2} , y = - \setminus \frac{1}{2}$

Therefore coordinates:

$\left(- \setminus \frac{3}{2} , - \setminus \frac{1}{2}\right)$