# math

posted by
**olive**
.

transform the ff. into the form f(x) = a(x - h)2 + k:

f(x)=x2-6x+4

f(x)=x2+4x+2

f(x)=3x2-5x+1

tnx a bunch!!!1

I'll do the third one for you, the first two are really easy

f(x)=3x^2-5x+1

=3(x^2 - 5/3 x ) + 1 factored out the 3 from first 2 terms

=3(x^2 - 5/3 x + 25/36 - 25/36) + 1 took half the coefficient of the x term, squared it, then added and subtracted it

= 3((x-5/6)^2 - 25/36) + 1 changed x^2 - 5/3 x + 25/36 to (x-5/6)^2, that is why it is called "completing the square"

=3(x-5/6)^2 - 25/12 + 1 multiplied by 3

=3(x-5/6)^2 - 13/12

of course a quicker way would be to find the x of the vertex using x=-b/(2a) = 5/(2*3) = 5/6 then subbingh that back into the equation to get y.

y = 3(25/36) - 5(5/6)+1

=25/12 - 25/6 + 1

=(25 - 50 + 12)/12

= -13/12

now form your equation

y = 3(x-5/6)^2 - 13/12