Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Answer 1

The equation is  

                        $y=2x^2-32x+56$.

Factoring 2, we get 
 
                        $y=2(x^2-16x+28)$.


We notice that $(x-8)^2=x^2-16x+64$, so $x^2-16x=(x-8)^2-64$.

Substituting in the previous equation, we have:

                       $y=2(x^2-16x+28)=2[(x-8)^2-64+28]=2[(x-8)^2-36]$, 

distributing 2 over the two terms inside the brackets, we finally get:

                                   $y=2(x-8)^2-72.$


This is a parabola opening upwards, since the coefficient of x^2 in the original equation is positive, and whose vertex is (8, -72), which is the lowest point of this parabola.


Answer:  $y=2(x-8)^2-72.$; x-coordinate of the minimum: 8.