From Standard to Vertex

In Section 1 you read a parabola's vertex from vertex form, $a{\left(x-h\right)}^{2}+k$. Completing the square is how you get there from standard form. The difference from solving is that an expression has no other side to add to, so you add the missing number and take it straight back off.

Take ${x}^{2}+8x+20$. The missing number for ${x}^{2}+8x$ is 16. Adding 16 and subtracting 16 changes nothing: ${x}^{2}+8x+16-16+20$. The first three terms are ${\left(x+4\right)}^{2}$, and -16 + 20 is 4, so the expression is ${\left(x+4\right)}^{2}+4$.

Now the vertex can be read: (-4, 4). The square is never negative, so the smallest the expression can be is 4, reached at x = -4.