# Thread: Finding vertex of parabola by completing the square?

1. ## Finding vertex of parabola by completing the square?

I must have been absent from school when we learned this because the only way I know how to find the vertex of a parabola is to graph it and calculate the maximum. Help is appreciated. Thanks in advance.

The equation of a parabola is y=x²-2x+7. Show how to determine the vertex of this parabola by completing the square. What is the vertex of the parabola?

2. Let's see how well I remember this one...

Ok, so we reorganize the equation:

$\displaystyle x^2 - 2x - y = -7$

So we want to complete the squares for the x terms. Take the -2x term and divide by 2 (so, -1) and square that (thus, 1). That is the number you need to complete the square. Do this:

$\displaystyle x^2 - 2x + 1 - y = -7 + 1$

Note that I added the +1 to the right side as well!

So now we can easily factor $\displaystyle x^2 - 2x + 1$, leaving us with:

$\displaystyle (x - 1)^2 - y = -6$.

Voila, square completed. You'll probably want to reorganize the equation a bit into the standard form:

$\displaystyle (x - 1)^2 = y - 6$

Can you take it from there?

3. Thanks! I'm thinking the vertex is (1,-6). The x-value is negated because we're taking it our of parentheses and the y-value remains the same.

4. Close!

Actually, the vertex is at (1, 6). Remember that the formula is:

$\displaystyle (x - h)^2 = 4p(y - k)$, where the vertex is (h, k).

In this case, 4p = 1, leaving just the (y - 6) on the right side for y = 6. Of course, with the (x - 1)^2, we get x = 1 for a vertex at (1, 6). Always good to graph it to make sure your solution is correct, I find.

5. I see! Thank you once again!

6. My pleasure, glad to help.