Since the coefficient of x^2 is 1, we can take the coefficient of x, the 20, divide it by 2 to get 10, and then square that number to get 100. Now add zero by adding and subtracting 100 to get:
x^2+20x=(x^2+20x+100)-100
But the expression x^2+20x+100 is a perfect square:
x^2+20x+100=(x+10)^2.
Hence,
x^2+20x=(x^2+20x+100)-100=(x+10)^2-100
One purpose of this procedure is to help you graph the function f(x)=x^2+20x. Since f(x)=(x+10)^2-100 as well, you can see that the minimum value of f(x) is -100 when x=-10 (note that the coefficient of (x+10)^2 is 1, which is positive, so the vertex is a minimum rather than a maximum). The vertex of the parabola that is the graph of f is at (x,y)=(-10,-100). The graph opens upward, and you can find other things to help you graph it, such as x and y-intercepts (the y-intercept is f(0)=0 and the x-intercepts are x=0 and x=-20 since f(x)=x^2+20x=x(x+20).
Can you use the method above to complete the square in the following general case? f(x)=x^2+bx+c? Give it a try and see if you can figure out where the vertex of the graph of f is after doing so (its coordinates depend on b and c).
