SOLUTION: please help me solve this equation in vertex form y=3x^2-6x+10

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: please help me solve this equation in vertex form y=3x^2-6x+10       Log On


   



Question 1074271: please help me solve this equation in vertex form y=3x^2-6x+10

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing it for you, here is one EXACTLY like the
one you have.  Only the numbers are changed.  Do yours step
by step as I have done this one.  If you have any trouble
just tell me in the email form below and I'll get back to 
you by email.

y=4x%5E2-16x%2B25

Factor 4 out of the first two terms

y=4%28x%5E2-4x%29%2B25

To complete the square:

1.  Begin with the coefficient of x.  That's -4
2.  Multiply it by 1/2.  That's (-4)(1/2) = -2
3.  Square it.  That's (-2)2 = +4
4.  Add it and subtract it inside the parentheses: We put in +4-4

y=4%28x%5E2-4x%2B4-4%29%2B25 

We factor the first three terms inside the parentheses:

y=4%28%28x-2%29%28x-2%29%5E%22%22-4%29%2B25

The two factors are the same so we write the product 
as (x-2)2.

y=4%28%28x-2%29%5E2-4%29%2B25

Now we remove the outer parentheses, by distributing,
leaving the (x-2)2 intact:

y=4%28x-2%29%5E2-16%2B25

Combine the numbers -16 and +25 as +9

y=4%28x-2%29%5E2%2B9    <--answer

Now do yours using this as a step by step guide.

Edwin