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Prove It $\displaystyle y = 3x^2 + 12x - 7$
Take out the common factor of 3
$\displaystyle y = 3[x^2 + 4x - \frac{7}{3}]$
Add a cleverly disguised 0... in other words, $\displaystyle \left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)$
$\displaystyle y = 3[x^2 + 4x + 2^2 - 2^2 - \frac{7}{3}]$
Can you see that $\displaystyle x^2 + 4x + 2^2$ is a perfect square?
$\displaystyle y = 3[(x + 2)^2 - \frac{19}{3}]$
Expand the common factor of 3 through the brackets.
$\displaystyle y = 3(x+2)^2 - 19$
This is now of the form $\displaystyle y = a(x-h)^2 + k$ where h and k are the x and y co-ordinates of the vertex.
So what is the vertex?