Find the vertex by completing the square.?

Question

f(x)= x^2-6x+5

Answer 1

f(x)=x^2-6x+5

f'(x)=2x-6

f'(x)=0 => x=3 => y= -4

Therefore vertex: (3,-4)

Answer 2

Rule of thumb: the magic number you need is going to be "half squared" of the term in front of the x. Let's keep this in mind when doing the question.

f(x) = x^2 - 6x + 5

Our term in front of the x is -6, so we want to take half of that and square it. Half of -6 is -3, squared is 9. So 9 is our magic number, and what we're going to do is add 9 and then subtracting it again (since 9 subtract 9 is zero, and adding zero doesn't change an equation, which is exactly what we want).

f(x) = x^2 - 6x + 9 + 5 - 9

Notice that I added and subtracted 9, and where I placed them. Now, the first three terms make up a perfect square, and this becomes

f(x) = (x-3)^2 + 5 - 9

Note how, upon changing the form, the term that goes in the brackets is half of the term in front of the x in the initial equation (half of -6 is -3). Now, we reduce even further to get

f(x) = (x-3)^2 - 4

To obtain the vertex from this, you take the *negative* of what's in the brackets as your x-coordinate, and the number outside of the brackets as your y coordinate. The negative of -3 is 3, and the term outside of the brackets is -4, so the coordinates of your vertex should be (3, -4).

Answer 3

Puggy is right but a little long-winded.

The way I like to think about it is you want to add and subract an amount (equalling zero, thus not changing the equation) such that you can 'absorb' part of it into your equation, making it a perfect square.

A simple example

x^-3x=5
adding and subtracting (-3/2)^2 doesn't change the equation:

x^2-3x+(-3/2)^2 -(-3/2)^2 = 5

(x-3/2)^2 = 5+9/4

Now you can take square roots and get the two solutions.

This is how the quadratic formula is derived.

Oh, and since you wanted the vertex, that's the midpoint between the roots.