Posted by Lena on .
The question is as followed:
y=3x^2 + 4x
This is how I solved it:
y=3x^2 + 4x
y= 3 (x^2 + 4/3x)
y= 3(x^2+4/3x +4/9) + 4/3
y= 3 (x+2/3)^2 + 4/3
The answer sheet said that x = 2/3 but wouldn't it equal 2/3 because there is a 3 before the bracket?

Find the minimum/maximum 
bobpursley,
y=3x^2 + 4x
y= 3 (x^2 + 4/3x)You erred here, it should be
y=3x^2 + 4x
y= 3 (x^2  4/3x ) 
Find the minimum/maximum 
Lena,
oh ok
thanks so much :)