Find the minimum value of the quadratic 3x^2-8x+1.

This is a question about completing the square. The first step involves taking the factor of 3 our of the expression to reach the correct form for completing the square to reach 3(x^2-(8/3)x)+1. Then, consider x^2-(8/3)x and complete the square of this expression. The coefficient of x is -(8/3) so half of that is -(4/3) so we get (x-(4/3))^2-(16/9). We now substitute this back into the expression before so we have 3x^2-8x+1=3(x^2-(8/3)x)+1=3((x-(4/3))^2-(16/9))+1=3(x-(4/3))^2-(16/3)+1=3(x-(4/3))^2-(13/3)and this is in the correct form for completing the square. To find the minimum value we simply have to notice that the smallest value the squared term can be is 0 so the minimum value of the whole expression is -(13/3).