complete the square

mathaddict · #2 $Old$ December 1st, 2008, 07:46 AM

Quote:

Originally Posted by william $View Post$

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

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

william · #3 $Old$ December 1st, 2008, 02:31 PM

Quote:

Originally Posted by mathaddict $View Post$

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

I don't understand what you did. What is the vertex?

Prove It · #4 $Old$ December 1st, 2008, 04:32 PM

Quote:

Originally Posted by william $View Post$

I don't understand what you did. What is the vertex?

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

Take out the common factor of 3

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

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

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

Can you see that $x^2 + 4x + 2^2$ is a perfect square?

$y = 3[(x + 2)^2 - \frac{19}{3}]$

Expand the common factor of 3 through the brackets.

$y = 3(x+2)^2 - 19$

This is now of the form $y = a(x-h)^2 + k$ where h and k are the x and y co-ordinates of the vertex.

So what is the vertex?

william · #5 $Old$ December 1st, 2008, 06:13 PM

Quote:

Originally Posted by Prove It $View Post$

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

Take out the common factor of 3

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

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

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

Can you see that $x^2 + 4x + 2^2$ is a perfect square?

$y = 3[(x + 2)^2 - \frac{19}{3}]$

Expand the common factor of 3 through the brackets.

$y = 3(x+2)^2 - 19$

This is now of the form $y = a(x-h)^2 + k$ where h and k are the x and y co-ordinates of the vertex.

So what is the vertex?

$V: (-2,-19)$