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11.25.2018: One Lesson of Math - Quadratic Function in Vertex Form, Part 1


Today's soundtrack is Tchaikovsky: Masters of Classical Music, Vol. 6.


This afternoon's lesson is "Analyzing Quadratic Functions of the Form y = a(x - p)² + q". Now, I've never heard of this form, so I did a Google search on it and found a video on Khan Academy which says that there are three forms of the quadratic equation: vertex form, factored form, and standard form. Apparently y = a(x - p)² + q is vertex form. Based on that information, I was able to find another video on Khan Academy called "Vertex Form Introduction." I'm going to peruse this video before attempting the lesson.

 

So, my initial impressions were correct: quadratic equations can be listed in three forms:

- Standard form, which is ax² + bx + c = 0

- Vertex form, which is y = a(x - p)² + q

- Factored form, which is a(x-p

(Side note: ironically, there isn't an agreed-upon standard for these names; some textbooks call the vertex form the standard form, and call what I referred to as the "standard form" above the expanded form.)


Vertex form is so called because it is the form in which it is easiest to find the vertex of the quadratic function. When we look at the vertex form y = a(x - p)² + q, we can find the vertex with very little calculation. Our x- coordinate will be the inverse of -p, and our y-coordinate will be the value of q.

 

Now, on to the textbook.


By merely looking at a graphed quadratic function, we can input its values into the vertex form framework and end up with a complete equation. We now know that the vertex form is y = a(x - p)² + q, and we know that the vertex's x-coordinate is the inverse of -p, and that our y-coordinate is q.


I'm going to use one of the examples from my textbook, Pearson's Pre-Calculus 11. The graph at the right is taken from the worktext.


We can start by identifying the graph's vertex - let's say, (2, 3). We input those values into our vertex form equation, giving us y = a(x - 3)² + 2. The missing values are a, x, and y. Now we can choose any other coordinate on the graph that intersects a point of our parabola. Let's choose (5, 4). The y-value of our chosen point will be inserted in for x in our equation, and we will put the x-coordinate into the y variable in our equation. a is left as a variable. We now have the equation 4 = a(5 - 3)² + 2. We then solve for a to find its value. In this case, we end up with a = ½.


In our final equation, we want to see x and y as variables, so that inserting a value into either one would give us its corresponding coordinate. So our final equation is y = ½ (x - 3)² + 2.

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