# Thread: complete the square

1. ## complete the square

$\displaystyle y=3x^2+12x-7$

2. ## Re :

Originally Posted by william
$\displaystyle y=3x^2+12x-7$

$\displaystyle 3x^2+12x-7=0$
$\displaystyle x^2+4x-7/3=0$
$\displaystyle (x+2)^2=7/3+(2)^2$

3. Originally Posted by mathaddict
$\displaystyle 3x^2+12x-7=0$
$\displaystyle x^2+4x-7/3=0$
$\displaystyle (x+2)^2=7/3+(2)^2$
I don't understand what you did. What is the vertex?

4. Originally Posted by william
I don't understand what you did. What is the vertex?
$\displaystyle y = 3x^2 + 12x - 7$

Take out the common factor of 3

$\displaystyle y = 3[x^2 + 4x - \frac{7}{3}]$

Add a cleverly disguised 0... in other words, $\displaystyle \left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$

$\displaystyle y = 3[x^2 + 4x + 2^2 - 2^2 - \frac{7}{3}]$

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?

5. Originally Posted by 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?
$\displaystyle V: (-2,-19)$