How do you write the standard form equation of the parabola # y = x^2 - 6x + 14#?

The equation is already in standard form, seeing as it is ordered from higher to lower exponent. However, to find the vertex form of the equation, we need to convert the #x#-containing terms into a factorable expression, namely the square of a difference. We need to remember that:

#(a-b)^2 = a^2 - 2ab + b^2#

and vice versa. Now, let's take a look at the terms containing an #x# in the equation:

#x^2 - 6x#

Not as much as a perfect square, but if #a = x#, then the second term will be #2xb#, so #b# is #3#. To complete the square, we require a "#+b^2#" in the end, so we need to add #9#.

However, we can't just do that, since that changes the equation. To "keep the balance", we need to also subtract #9#. This becomes:

#y = x^2 - 6x +color(red)(9) + 14 -color(red)(9)#.

Using the expression above, factor the first three terms into a difference of squares, then do the subtraction #14 - 9 = 5#:

#y = (x-3)^2 + 5#

In general, the vertex form helps us find the vertex of a parabola. If the vertex form is

#y = a(x-b)^2 + c#, where #a# is the "coefficient" of #x^2# in the original equation, then the vertex is the point #(b,c)#.