You may want to review equations of parabolas and completing the squarefirst.

Our new form for a quadratic function is useful in other ways
too. First of all, the graph of a quadratic function looks
somewhat like the letter U. However, it can point in any
direction. To find out which, look at the number in front of the
part with the *x*. (For example, if it were *y* =
2(*x* −5)^{2} + 6, we'd look at the 2.) Since 2
is a positive number, the U opens upwards (like the actual letter
U). If we had a negative number there instead, then the U would
open downwards (like ∩). If the variable were *y* instead
of *x*, then it would open to the right (sort of like ⊂)
if the number were positive and to the left (sort of like ⊃)
if it were negative.

If you understand this, then it's much clearer what the point of
the vertex is. The vertex is just the tip of the U (the bottom, if
it's right side up). It has an *x*-coordinate, which is what
you've already learned to find (5 in our example), but it also has
a *y*-coordinate. We can read that off from our equation,
since it's just the constant added in at the end. In this case,
that's 6, so the vertex is the point (5, 6). (Remember, a point is
given by an ordered pair of numbers.

Remember that, when you write an ordered pair, the first number
is *x* and the second is *y*. So, if the equation
were *x* = 2(*y* −5)^{2} + 6, the vertex
would be (6, 5), since we always write the *x* first.

Suppose we have a quadratic function in the old form,
like *y* = −2*x*^{2} − 4*x* +
1. We can get it into the new form by completing the square.
First, we factor the −2 out of the first two terms,
getting *y* = −2(*x*^{2} + 2*x*) + 1.
Then the part in the parentheses is the beginning
of *x*^{2} + 2*x* + 1, which is the square
of *x* + 1, so it becomes *y* = −2[(*x* +
1)^{2} −1] + 1, which simplifies to *y* =
−2(*x* + 1)^{2} + 2 + 1, or *y* =
−2(*x* + 1)^{2} + 3. So the graph opens
downwards, and its vertex is (−1, 3).