## Tuesday, February 19, 2013

### More on quadratic equations and parabolas

This week my studies focused on the vertex of the graph of a quadratic equations. The vertex of a parabola is the point in which the graph changes direction. Imagine the highest point of a roller coaster which you climb before plummeting downwards. That is the vertex. Now imagine a slope between two mountains, the vertex would be the point where you stop coming down one of the mountains and start coming up the other. We can find the vertex of a quadratic equation in standard form by transforming it to vertex form.

The way the vertex transformation works is the following: If you can get your quadratic equation in the form of y =a(x-h)² + k, remembering that (x-h) is equal to (x+(-h)), then (h,k) is the vertex of that equation. If a is positive, then the vertex is the minimum point the parabola reaches before going up, if a is negative then the vertex is the maximum point the parabola reaches before going down.

The vertex is an important point to know since if someone gives you the vertex and another point in the parabola, you can get the standard equation for it. For example, if I know the vertex of my parabola is (3,-4) and that point (2,-3) is in the parabola we can use algebra to get the equation of it.

y =a(x-h)² + k
y =a(x-3)² + (-4) substituting the given vertex into the equation.
-3 =a(2-3)² + (-4) substituting the (x,y) point given we the solve for a.
1=a or a=1
Therefore the equation of the parabola is (since a=1, we omit it):
y =(x-3)² - 4
Or (x-3)(x-3) -4
which gives us the standard equation x²-6x+5.
Now that is cool.

It seems most of my time while going through pre-calculus in order to get to calculus has been spent on quadratic equations. I did not remember there was so much to learn from them. Whenever I am sure I am done with parabolas something else comes up. Ad that something else is amazing to understand.

Right now when I look at x2-6x+5, I can:
• Look at the standard form of that quadratic equation and tell at what point the graph passes the y axis (y intercept) and know whether the graph is concave up (like a cup) or concave down (like and umbrella)l
• Solve quadratic equations using the quadratic formula and thus getting the points were the graph crosses the x axis (x intercepts).
• Use the discriminant in the quadratic formula to know wether the equation has rational, irrational or imaginary solutions as well as the number of x intercepts.
• Use factoring (when possible) to get the same results as with the quadratic equation.
• Transform that standard form to the vertex form (y = a(x-h)² + k) to find the vertex or the point where the parabola changing direction (its maximum or minimum). I will talk about this form later in the post.
And I can graph it since I know these points:
Y intercept is 5 so (0,5)
2 X intercepts 1 and 5 so (1,0) and (5,0)
The vertex is (3,-4)

Let's graph it:

Notice I added a new point at the top right corner, (6,5). I knew that point should be there since a parabola is simetrical across its vertex. In other words, one side is a mirror image of the other. And since point (0,5) was on the left side, then point (6,5) must exist in the right. If you want to be sure, just plug 6 in the formula substituting it for x. I'll wait until you do so.

Going back to the vertex transformation, that transformation got me to think that someone had to discover this relationship between quadratic equations and their graphs through serious and hard study. I must remind myself that all these formulas, equivalencies, and transformations I am now given, where not always available. It is easy to forget that someone put a lot of effort coming up with these "shortcuts" we now use is class. I thank immensely all those mathematical scholars who made this learning possible.

If you liked this quadratic stuff as much as I did let me know.