In search of the parabola's vertex...

Today's lesson was a combination of a review from other classes and new things that goes a little more in-depth on parabolas. For example, this is what we put in our Math Dictionary, but due to a shortened period of time (we had an assembly and before that, a quiz), this is all we could write. (I'm a luckier scribe! =P) "I won't be posting what we reviewed because all the things that we reviewed are posted after this post {scroll down}."

There are 3 different forms of equations:

standard form: f(x) = a(x - h)² + k
factored form: (x + 4)(x - 2)
generalized form: x² + 2x + 7

The generalized form can give us a lot of info about the parabolas. Factor it, you can find the roots, then you can find the Axis of Symmetry which means you can find the vertex.

If a quadratic function is factorable (is not prime; ie. has zeros), then it can be written in factored form as the product of two binomials.

Ex.

f(x) = x² + 2x + 8
= (x + 4)(x - 2)

[it looks like all that factoring we did in gr. 9 and gr. 10 all pays out in gr.11 =)]

From the equation, we can get two binomials. If we try and make one of the binomials equal 0, then we would know the x-intercepts in the parabola which are its roots.

(Tip: x-intercepts have a y-coordinate of 0)

like so...

In (x + 4), if x = -4, then...

(x + 4)(x - 2) = [(-4) + 4][(-4) -2]

= (0)(-6)

= 0 <-- this means that the y-coordinate is 0 because it's the x-intercept

In (x - 2), if x = 2, then...

(x + 4)(x - 2) = [(2) + 4][(2) - 2]
= (6)(0)
= 0 <-- this means that the y-coordinate is 0 because it's the x-intercept

(insert diagram of the parabola of the equation here)

As you can see in the above diagram, the x-intercepts are the roots of the parabola. If that is true (and it is), the vertex of the parabola lies in between the two roots which also happens to be lying on the Axis of Symmetry.

Flashback...

Axis of Symmetry: The line (of reflection) through which one half of the parabola "folds" onto the other half...

To find the midpoint of the two roots, you add them up and divide by two like so:

(-4 + 2)/2 = -1 <-- This -1 is the x-coordinate of the vertex. (If you look in the previous diagram that the x-coordinate of the vertex is -1) :O

x-coordinate of vertex = Axis of Symmetry

Remember that equation from the beginning of the lesson?:

f(x) = x² + 2x + 8
= (x + 4)(x - 2)

Well, we now know that the x in the f(x) is -1 because if we label that -1 as x (because its an x-coordinate) and insert that into the f(x) in the equation...

f(-1) = [(-1) + 4][(-1) - 2]
= (3)(-3)
= -9

...we get the y-coordinate of the vertex. :D

So the ordered pair of the vertex is (-1, -9). (Just like the vertex of the parabola in the diagram, they have the same coordinates for the vertex)

So, from today's lesson, we also found out that "to factor" really means "to find where the roots are in the parabolas."

But when an equation turns out to be prime....

Ex.

x² + 2x + 7 <-- we cannot factor this trinomial so we consider it as prime

...then it will not factor out nicely such as the previous trinomial where we got the (x + 4)(x - 2). Well we do know that this equation is a parabola so it must be able to be factored.

(insert pic of x² + 2x + 7 parabola here)

In this case, it's true. THIS PRIME TRINOMIAL CAN BE FACTORED, but the catch is...it cannot be factored nicely into whole numbers which we will use radicals instead.

Flashback...

Radicals are those numbers with that thingamajiggy called "radical sign".

Ex. √ 4 (the square root of 4 is 2)

That topic on how to factor these "unfactorable" trinomials, will be covered in the distant future.

~ and that's about it for today's lesson and we have no homework. =D
...well math homework to be exact... =(

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oh, and one more thing, I pass the baton to cherrie