Thread: complete the square
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Old December 1st, 2008, 05:32 PM
Prove It Prove It is offline
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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?
Last edited by Prove It; December 2nd, 2008 at 02:53 AM.
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