f(x)=3x^2+5x-4

please go through the steps to get to this: f(x)=3(x+5/6)^2- 73/12

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- Oct 23rd 2007, 03:17 PMbilbobagginsconvert to vertex form.
f(x)=3x^2+5x-4

please go through the steps to get to this: f(x)=3(x+5/6)^2- 73/12 - Oct 23rd 2007, 04:38 PMticbol
Hello, bilbobaggins!

For your taking care of the Ring for 60 years, I offer my humble help.

f(x) = 3x^2 +5x -4

You want that, Sir, in its "vertex form", or in the form f(x) = a(x-h)^2 +k, where (h,k) i its vertex.

That is done by "completing the square"

f(x) = [3x^2 +5x] -4

f(x) = 3[x^2 +(5/3)x] -4

f(x) = 3[x^2 +(5/3)x +(5/6)^2 -(5/6)^2] -4

f(x) = 3[(x +(5/3)x + (5/6))^2] -3(5/6)^2 -4

f(x) = 3[(x +5/6)^2] -3(25/36) -4

f(x) = 3[(x +5/6)^2] -75/36 -4

f(x) = 3[(x +5/6)^2] -(75 +36*4)/36

f(x) = 3[(x +5/6)^2] -219/36 -----------that's it.

Sorry, Sir, but 73 is way too low for 219.