## Lesson Overview

In this math lesson students will learn:

- How to solve quadratic equations using the zeroes form (y = C(x – z1)(x-z3)) and the vertex form (y=A(x – h)2 + k).
- How to identify which form is more useful for solving quadratic equations, based on whether they are provided with x-intercepts, y-intercepts, the vertex point of a graph in (x, y) form, and another point on the graph.
- How to determine whether or not equations are equivalent and how to check equations algebraically against any given data.

Students will also:

- Listen to a lecture presented by the math teacher in which he or she explains how one can use the zeroes form and the vertex form to solve quadratic equations.
- Follow and solve sample equations provided by the math teacher during the course of the lecture.
- Learn and review the following vocabulary words:
**zeroes form, vertex form, x-intercept, y-intercept, parabola, vertex, scaling constant, general form equation.** - Ask questions during the lecture to gain clarification of unfamiliar concepts.

## Learning Activities

1) The math teacher should begin the lesson with a review of using the general form (y=ax^{2} + bx +c) to solve quadratic equations.

2) Next, the math teacher should explain the use of the zeroes form y = C(x – z_{1})(x-z_{3}) to solve quadratic equations, using a vertical parabola with the x-intercepts of x = -3 and x = 1. Plugging in these x-intercepts, the equation looks like this: y=C(x – (-3))(x – 1), which can then be rewritten as y=C(x + 3)(x – 1). To solve for C, which is the scaling constant, another point must be provided. In this case, the example is (0, -12). When this point is plugged into the equation, it looks like this: -12 = C(0 + 3)(0 + 1), which translates to -12 = 3C. Solving for C gives us 4, which provides the final equation: y = 4(x + 3)(x – 1).

3) The teacher should next move on to the use of the vertex form (y=A(x – h)^{2} + k), which can be used when the vertex (e.g., the highest or lowest point on a parabola) is given, along with a third point on the graph. To solve this equation, the instructor should use the vertex (-1, -16), which is (h, k), along with a third point (0, -12) to find A, which is the scaling constant. With these three points, the equation now looks like this: -12 = A(0 + 1)^{2} – 16, which results in A = 4. The final equation looks like this: y = 4(x + 1)^{2} – 16.

4) The next part of the lecture will cover determining whether the equations are equivalent, so make sure there are no rounding errors. To do this, the zeroes form equation and the vertex form equation need to be written out in general form using exact arithmetic. Each form is then checked to see if they have the same general form; if they do, they are equivalent.

5) The last part of the lecture will briefly cover algebraically checking an equation by plugging the coordinates of a point from the parabola into the equations and making sure both sides are equal.

## References

*Image Credit: Vertex-Oleg Alexandrov/Wikimedia Commons/Public Domain*