Question 6520
I'm wondering where you learned that formula y - a = m(x - b) as the vertex-point form for a quadratic function. I'm also particularly curious as to what the m actually is, and I don't think it's slope in this case.


A quadratic function can be written in the form y = a(x - h)^2 + k where (h,k) is the vertex of the parabola. We're gonna go through a long process called completing the square to get to the answer (2,-4) for our example.


The equation as thrown to you is {{{ x^2 - 4x - 2y - 4 = 0 }}}. First of all, we need to put this in standard quadratic form so that we can write it as "y = ".


{{{ x^2 - 4x - 4 = 2y }}} <---- Moved the 2y to the other side of equation


{{{ y = (1/2)x^2 - 2x - 2}}} <---- Divided both sides by 2 and put the y in front.


We haven't "completed the square yet." This is a tricky process, and we must go through some basics first before going on.


First up, let's say that we'll begin with a number a. If we say a + 3 - 3, we really didn't change the value of a. This may seem like a "duh" but we'll need to use this trick later on.


Say that we have a*(b) to begin with. Suppose that we are to add a "+ 3" to the b inside the parentheses, so that a(b + 3). Now, how do we undo that so that we'll get back to the original a(b)? Since a(b + 3) = ab + 3a, we'll need to subtract the 3a from ab + 3a to get back to a(b). In other words a(b) = ab + 3a - 3a = a(b + 3) - 3a. This is just a step above the trick mentioned in the first paragraph. This is the trick you'll be using for this example because you have an a-value that isn't 1.


The second thing we need to establish is the recognition of trinomial squares and backwards thinking with them. Say that we have (x + 3)(x + 3). We know that that's equal to x^2 + 6x + 9. When we see a quadratic expression in the form ax^2 + bx + c, we must know how to tell whether it's a perfect trinomial square, aka, whether it can be written as (x + n)(x + n). The way to do that is to take the b term, in this case, 6. Divide that by two (which brings it to 3) and square it, which brings it to 9. Does the result equal the c term of the expression? If yes, then the expression is a trinomial square. We will use this fact heavily in completing the square.


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On with the game.

{{{ y = (1/2)x^2 - 2x - 2 }}} <----- Start with quadratic function in standard form.


{{{ y = (1/2)(x^2 - 4x) - 2 }}} <----- We factored 1/2 from the first two terms and left the -2 untouched. (If you performed distributive property on this equation, you'll end up with {{{ y = (1/2)x^2 - 2x - 2 }}}. The "increase" of the -2x to the -4x confuses most people because factoring usually cuts down a number, but we're factoring out a fraction less than 1, so the effect is opposite)


Hang on because here's where the tricks come into play. We're going to first force the {{{(x^2 - 4x)}}} to be a perfect trinomial square. The b term in that expression would be the -4. Cut it in half, you'll get a -2. KEEP THIS NUMBER HANDY BECAUSE YOU'LL USE IT. Now take that -2 and square it. That brings you to +4. KEEP THIS NUMBER HANDY AS WELL!!! Now, the {{{(x^2 - 4x) }}} term turned to {{{ (x^2 - 4x + 4) }}} which is a "forced" perfect trinomial square. The process of taking the b term, halving it, and squaring the result guarantees a perfect trinomial square.


So far, our working function is 


{{{ y = (1/2)(x^2 - 4x + 4) - 2 }}} but we're not done yet. 

{{{ y = (1/2)(x^2 - 4x + 4) - 2 -4(1/2) }}} <----Remember the trick a(b + c) - ac = ab. Since we added a "+ 4" inside the parentheses, we must take it out again to preserve balance. Since the +4 really is being multiplied by the 1/2, we'll have to take out 4(1/2) = 2.


{{{ y = (1/2)(x^2 - 4x + 4) - 4 }}} <---- We're almost done.


Remember when we were to make note of the b term when we force-created a trinomial square? It was that -2. When we rewrite the equation just above with the -2, it becomes


{{{ y = (1/2)(x - 2)^2 - 4 }}} <---- The move from (x^2 - 4x + 4) to the (x - 2)^2 is backwards FOILING. You don't need to worry because that step where you (take the b term, divide it by two, and square the result) guarantees that the whole thing will work.


Now, the quadratic equation is in a form where you can easily pull out the vertex of the parabola. That is (2, -4). (Once in this form, the vertex's x-coordinate is the number after the x variable inside the parentheses WITH the OPPOSITE sign. The y-coordinate is whatever the last constant is, and YES, preserve the sign!)