GeoGebra Tutorial: Graphs and Sliders Part 2

In my post Tutorial 4: Graphs and Sliders, we learned how to use the slider in investigating graphs of the form y = mx + b. In this tutorial, we are going to use sliders to investigate the graph of the fromy = a(x – h)2+ k, where a, h and k can be any real number.  We will first input the values of a, h and k in the input box before creating the sliders.

To do this, first we are going to assign temporary values for a, h and k in the input box, then create a slider for each of them. After creating a slider, we enter the equation y = a(x – h)2+ k in the input box to graph our function.  You may want to look at our expected output here.


    1. Open GeoGebra.
    2. Enter the following equations in the input box: a = 1, h = 1 and h = 1 in the Input box and press the ENTER key on your keyboard after typing each equation.  Observe that the equations appear in the Free Objects section of the Algebra window.

Figure 1 – Equations are entered in GeoGebra using the Input box.

    1. Right click each equation and click Show Object. Notice that the sliders appear in your drawing box.

Figure 2 – Sliders are made by right-clicking each equation and pressing the Show Object option in the context menu.

    1. Type the equation y = a*(x – h)^2 + k, the press the ENTER key. If you have typed the equation correctly, a graph should appear in your drawing pad.

Q1: Click the Move button and move the small circle on your slider. What do you observe?
Q2: What are the effects of the parameters a, h and k to the graph of the function y = a*(x – h)^2 + k?

Why is any number raised to 0 equals 1?

Why is any number raised to 0 equals 1?

If we raise a number to an exponent, we are multiplying it by itself a certain number of times.  For example, 3^4 means you have to multiply 3 by itself 4 times.   In exponential notation, we call 3 the base and 4 the exponent.

Shown below are examples of exponential expressions and their expansion.

2^3 = 2 \cdot 2 \cdot 2

4^7 = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\cdot 4\cdot4

x^4 = x \cdot x \cdot x \cdot x

(a + b)^3 = (a+ b)(a+b)(a+b)

x^m = x \cdot x \cdot x \cdot \ldots \cdot x \cdot x (Multiply x by itself, m times)

The\ldots symbol means “and so on.” It represents x’s that are missing.  It is convenient to use the said symbol for large values of m.

Multiplying Expressions with Exponents

If we want to multiply expressions with the same base, let us see what happens. For example, what will happen if we multiply 2^3 and 2^5?

From above, 2^3 = 2 \cdot 2 \cdot 2 and 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2. Multiplying the two expressions, we have

2^3 \cdot 2^5 = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^8

From our computation, we can conclude that if we multiply to expressions with the same base, we have just have to add their exponents (Can you see why?). That is, for expressions x^m and x^n,

x^m \cdot x^n = x^{m+n} (*)

Q1: What if the base of the two expressions are not the same? Will our formula above still apply?

Dividing Expressions with Exponents

What about dividing expressions with exponents? Suppose, we want to divide 2^5  by 2^3.

We know that 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 and 2^3 = 2 \cdot 2 \cdot 2. Dividing the two expressions, we have

\displaystyle \frac{2^5}{2^3}= \displaystyle\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2} = 2^2

Since, three 2‘s are canceled out, we can therefore conclude that in dividing two expressions with the same base, we just have to subtract their exponents. That is, for expressions x^m and x^n,

\displaystyle\frac{x^m}{x^n} = x^{m-n}. (**)

What happens if the exponent of the denominator is larger? For example, \displaystyle \frac{2^2}{2^7}?

From (**), \displaystyle\frac{2^2}{2^7} = 2^{2-7} = 2^{-5}.

Now, let us compare this result when we expand our expression:

\displaystyle \frac{2^2}{2^7}= \displaystyle\frac{2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{1}{2^5}

Our observation tells us that, 2^{-5} = \displaystyle\frac{1}{2^5}. Therefore, x^{-m}= \displaystyle\frac{1}{x^m}.

Q2: In general, what is the value of \frac{x^m}{x^n} if m < n?

What if the exponents of the numerator are equal? For instance, \frac{2^5}{2^5}. This is practically dividing the same number, so obviously the answer is 1. However, we can also use our conclusion above.

From (**), \displaystyle\frac{2^5}{2^5} = 2^{5-5} = 2^0 = 1


Here we observe that raisingx (or any expression) to 0  means that the number of factors in the numerator and the number of factors in the denominator is the same. Therefore, x^0 can be expressed as x^{m-m} for any value of m. But from (**), x^{m-m} is equivalent to \displaystyle\frac{x^m}{x^m} = 1

Therefore, any number raised to 0 (with the exception of 0) equals 1.

Why do we reverse/flip the inequality sign?

You have probably remembered in Algebra that if we multiply an inequality by a negative number, then the inequality sign should be flipped or reversed. For example, if we want to find the solution of the inequality -\frac{1}{2}x > 8, we multiply both sides by -2  and reverse the greater than sign giving us x < -16. Now, why did the > sign became <?

If we generalize the statements above, suppose we have two numbers, say, a and b such that a > b, if we multiply them to a negative number c, instead of having  ac > bc,  the answer should be ac < bc.

Before we proceed with our discussion, let us first remember 2 basic concepts we have learned in elementary mathematics:

  1. The number line is arranged in such a way that the negative numbers are at the left hand side of 0 and the positive numbers are at its right hand side such as shown in Figure 1.
  2. If we have 2 numbers a and b, then  a > b if a is at the right of b on the number line. For example, in Figure 1, 2 > -1 since 2 is at the right of -1.
Number line

Figure 1 – The number line

For specific values, let’s choose a = 2 and b = -1 as shown in the diagram above and choose c = -1. Note that we will  just use these values for discussion purposes, but we may take any values. It would help, if we think of a and b as two points on the number line with a as a blue point on the right b, a red point.

And note that before multiplying with a negative number, VALUE OF BLUE POINT > VALUE OF RED POINT.

Since a and b are variables, we need to multiply all the numbers on the number line by -1. This is to ensure that whatever values we choose for a and b, we multiply them by -1. If we multiply every number on the number line by -1, the geometric consequence would be a number line with negative numbers on the right hand side of 0, and positive numbers at the left hand side of 0 as shown in Figure 2.


Figure 2 – Afer multiplying all numbers on the number line by -1

But negative numbers should be at the left hand side of 0 so we reverse its position by rotating it 180 degrees from any point of rotation (for example, 0).  The resulting figure is shown in Figure 3.

Figure 2 - All numbers in the number line were multiplied by -1

Notice that the blue and red points changed order and that the blue point is now at the left of the red point. Therefore, VALUE OF BLUE POINTVALUE OF RED POINT. That is, why the inequality sign was reversed.

Summarizing, multiplying an inequality by a negative number is the same as reversing their order on the number line. That is, if a, b and c are real numbers, a > b and c<0, then ac < bc.

Our summary above is actually a mathematical theorem. The proof of this is shown below. It is a very easy proof, so, I suppose, that you would be able to understand it.

Theorem: If a, b and c are real numbers, with a > b and c<0, then ac < bc.


Subtracting b from both sides, we have a - b>0.

Now, a - b>0 means a - b is positive.

Since c is negative, therefore, c(a - b) is negative (negative multiplied by positive is negative)

Since c(a - b) is negative, therefore, c(a - b) < 0.

Distributing c, we have ac - bc < 0.

Adding bc to both sides, we have ac < bc which is what we want to show .\blacksquare

Related Posts Plugin for WordPress, Blogger...
1 29 30 31