The M-Lectures: PARABOLAS and QUADRATIC EQUATIONS.

Polynomials

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How to sketch it. (Graphing Parabolas)

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Another look at deriving the Quadratic Formula

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Subtract C from both sides.

Solve for x. Cancel the a from the first term.

Complete the square trinomial on the left side.

Factor the left side perfect square trinomial. Rearrange the terms on the right side.

Find the LCM and subtract the fraction on the right side.

Multiply and rearrange the terms in the second fraction on the right hand side.

Take the square root of each side.

Simplify the square root of the LCM.

Leave the x by itself and you are done.

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Deriving the quadratic equation 6. (last)

Note: notice that c below was given to be the initial velocity, this is wrong. C is in fact the height at which the object is either thrown or catapulted. I will change this picture later.

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Deriving the quadratic formula 5.

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Deriving the quadratic formula 4.

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Deriving the quadratic formula 3.

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Deriving the quadratic formula 2.

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DERIVING THE QUADRATIC FORMULA 1.

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PARABOLAS

Some parabolas open up while others open down. If it opens upwards, the parabola has a minimum ("hold water.") If it opens downwards, it would "spill water," and it has a maximum point. A parabola facing up, is called "concave up." A parabola facing down is called: "concave down."

Graph the function: Y = - x exp2 + 4x + 3.

Step 1. Find the x-and the y-coordinate of the vertex of the parabola before constructing a table of values to graph it. The maximum is at (2, 7) this point is the vertex or maximum. Use 1, 0, 3 and 4 as the domain to find the corresponding range values. The formula to find the x coordinate of the vertex is : x = -b/2a. The axis of symmetry passes through the vertex. Note: that there are other ways to find the vertex using another vertex formula which is discussed in more advance math courses. You might want to do research on this. At the moment I am only covering the basic facts. Later in these series I would dedicate sometime to that section. For the moment, some basic facts are enough.

Step 2. Graph the parabola.

Use the quadratic formula to show the roots of the quadratic equation are real numbers. The quadratic formula show that it is, if not impossible, to graph its roots--since the real are infinite in nature and some of the impossible to sketch or graph with any specific certainty. Later we will see how easy it becomes, to graph certain types of quadratic functions. It is a fact of life, one cannot graph some points on a number line ever as their existence belong in Noumena.

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Computing the axis of symmetry before constructing a table to graph a quadratic equation.

1. The x-coordinate of the vertex is given by x = -b/2a.
2. Now we are ready to construct a table of values.

3. Computing the corresponding values for the y-coordinates.
4. Graphing the parabola.
5. Since a is positive the parabola has a minimum, i.e. It opens upwards.

6. In this case the roots are imaginary, since square roots of negatives give imaginary or complex results. Only points in the x-axis, give real roots.

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Fundamental Question:

Standard Form

Values of a, b, and c

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1. Identify the values of a, b, and c.

2. Write the two real roots.

3. We can write our result as two equations equal to 0. Only a mechanical move.

4. This result shows that not all quadratic equations can be graphed using real roots. Sometimes it is necessary to make a table of values.

Making a table of values should not be a guessing gave. We need to make sure we use the axis of symmetry to determine what values to use for the x-coordinates. The axis of symmetry gives us the location where in the coordinate plane to graph our parabola. Below is I started with x = 0 only because I knew in advance the x-coordinate 0 was mighty close to the axis of symmetry of this particular graph.

Simple computations. You must know how to do this well. Practice on your own.

The x-coordinate of the vertex is 1. The y-coordinate is -7. We know these values from our table. The axis of symmetry is not graphed here, you must use your imagination. The formula to find the x-coordinate of the vertex was used here. This confirms that our table of values is correct. In the future we must compute the x-coordinate of the vertex first to help us guide our table of values. Notice that this parabola has a minimum value and since a is positive, it opens upwards.

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The graph of a linear function is a line.

The graph of a quadratic function is a parabola.

If a is negative in the standard form in a quadratic function, the parabola opens downwards. Notice the vertex of this parabola is a maximum value at the origin.

X-squared? Why?

Quadratus mean?