I really unsure how to start with this, any help would be much appreciated

Find the vertex form of the quadratic function y = 3x^2 + 5x + 9

Find the vertex of the parabola which is the graph of the quadratic function y = -2x^2 + 4x + 6

2. Vertex form is when you complete the square.

$y = 3x^2 + 5x + 9$

$\Rightarrow y = 3(x^2 + \frac{5}{3}x + 3)$

$\Rightarrow y = 3[(x+\frac{5}{6})^2 - (\frac{5}{6})^2 + 3]$

$\Rightarrow y = 3[(x+\frac{5}{6})^2 - \frac{25}{36} + 3]$

$\Rightarrow y = 3[(x+\frac{5}{6})^2 + \frac{83}{36}]\quad$ (because 3 = 108/36)

$\Rightarrow y = 3(x+\frac{5}{6})^2 + \frac{83}{12}$

For the above example, the vertex is at $x = -\frac{5}{6}$. It is the number which makes the squared bracket = 0. For Q2 do the same as above then you can find the vertex.