# What is the axis of symmetry and vertex for the graph y=2x^2+6x+4?

Feb 26, 2017

Vertex is $\left(- \frac{1}{2} , - \frac{3}{2}\right)$ and axis of symmetry is $x + \frac{3}{2} = 0$

#### Explanation:

Let us convert the function to vertex form i.e. $y = a {\left(x - h\right)}^{2} + k$,

which gives vertex as $\left(h , k\right)$ and axis of symmetry as $x = h$

As $y = 2 {x}^{2} + 6 x + 4$, we first take out $2$ and make complete square for $x$.

$y = 2 {x}^{2} + 6 x + 4$

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

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

= $2 {\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{2} + 4$

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

Hence, vertex is $\left(- \frac{1}{2} , - \frac{3}{2}\right)$ and axis of symmetry is $x + \frac{3}{2} = 0$
graph{2x^2+6x+4 [-7.08, 2.92, -1.58, 3.42]}