• ## Top Posts

 quadratic equation c… on Modeling Data With Quadratic… Lemuel on Quadratic Functions and Their… Arizona Bayfield on How To Simplify Rational … Suemac on Proving Lines Parallel with Tr… Mr. Pi on Subsets of Real Numbers

## The Diagonals of a Rhombus are Perpendicular

This post is dedicated to proving the diagonals of a rhombus are perpendicular. To complete this or any proof, it is good to make a plan. In this case, I am going to establish that the two triangles on the top of the rhombus are congruent. This is seen in steps 2 through 4 of the proof below. Next, I show that angles AEB and CEB right angles, which is modeled in steps 6 through 8. That is enough to state the diagonals of a rhombus are perpendicular.

I encourage you to click on the image to see it at full size. It looks much better at full size. I scanned the piece of notebook paper that I work this proof out on and messed around in photoshop.

Proof - Diagonals of a Rhombus are Perpendicular

## How to Write Indirect Proofs – Exterior Angle Inequality Theorem

A youtube viewer of mine requested a video on how to write an indirect proof. Before making the video, I thought it would be good to write a blog post about this topic before I make my video, because an indirect proof is best written as a paragraph proof. There are three steps to writing an indirect proof.

1. Assume that the conclusion is false, by negating the prove statement.
2. Establish that the assumption in step #1 leads to a contradiction of some fact i.e. definition, postulate, corollary or theorem.
3. State the assumption must be false, thus, the conclusion or prove statement is true.

Steps 1 and 2 involve all of the thought and memory skills and can be discussed separately. Do not make this harder than it is. Step 1 involves writing the negation of a statement. Step 2 requires you to pull on your knowledge of geometric definitions, postulates, theorems and corollaries to recognize the contradiction between a known geometric fact and the assumption in step 1. Step 3 involves stating the obvious: Since the assumption is false, the prove statement must be true. If you are confused, check out the examples.

Example 1 – Prove the Exterior Angle Inequality Theorem with Indirect Proof

Given: $\angle{1}$ is an exterior angle of $\Delta{ABC}$

Prove: $m\angle{1}>m\angle{4}$

Figure 1 - Indirect Proof Diagram

Step 1 – Assume that $m\angle{1} \not> m\angle{4}$,  this means that $m\angle{1} \leq m\angle{4}$.

Step 2 – We need to establish that $m\angle{1} \leq m\angle{4}$ contradicts a mathematical fact.

$m\angle{1} \leq m\angle{4}$ gives two different situations that need to be tested:

$m \angle{1}= \angle{4}$ or $m \angle{1}.

$m \angle{1}= \angle{4}$

By the Exterior Angle Theorem, $m \angle{3}+m \angle{4}=m \angle{1}$ and using substitution, $m \angle{1}+m \angle{4}=m \angle{1}$. Subtracting $m \angle {1}$ from both sides gives $m \angle{4}=0$. This contradicts the fact an angle must have a measure greater than 0.

$m \angle{1}

By the Exterior Angle Theorem, $m \angle{3}+m \angle{4}=m \angle{1}.$ Angles must have a positive measure, the definition means $m \angle{1}> m \angle{3}$ and $m \angle{1}>m \angle {4}$.

Step 3 – In each instance, the assumption from step 1 is contradicted of a know mathematical fact. Thus, the assumption that $m\angle{1} \leq m\angle{4}$ is false. So, the original prove statement,  $m\angle{1}>m\angle{4}$ , must be true.

## Using Properties of Parabolas to Graph a Parabola Problem 2

As I discussed in an earlier post on graphing a parabola, the following information allows to easily graph a parabola in standard form $f(x)=ax^2+bx+c$.

Properties of a Parabola or a Quadratic Function in Standard Form

Given that $f(x)=ax^2+bx+c$ is a parabola when $a\neq0.$

1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
2. The axis of symmetry is given by the line $-\frac{b}{2a}$.
3. The vertex is given by the ordered pair $(-\frac{b}{2a},f(-\frac{b}{2a}))$
4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

In my first post the equation did not have a linear term and was in the form $f(x)=ax^2+ c$. The vertex could be found using (0,c) and the axis of symmetry (AOS) of x = 0. In terms of the coordinate plane, the y-axis is the AOS.

In this post, we change gears and will be graphing a parabola that is given in standard form $f(x)=ax^2+bx+c$. The point of this work is to sketch a graph using the core concepts behind the parabolic relation. Remember, because a parabola is symmetric every point except the vertex has a mirror image. Once you find the AOS and vertex, you will have to use the same process to find another point and then its mirror image. Let’s take this to the example.

Example Graphing a Quadratic Equation in Standard Form

Graph $f(x)=2x^2+4x-3$

Looking at the properties above in order, first we know this graph will open upwards, this is due to the fact the coefficient on the squared term is positive or a>0. The AOS is found by using the ratio $-\frac{b}{2a}=-\frac{4}{2(2)}=-1$ and can be written as x = -1. If you are trying to create the graph at the bottom of this post, you should draw a vertical dashed line at x = -1, because this represent the AOS.

Next we find the vertex. Since the equation or function is given as $f(x)=2x^2+4x-3$ and x = -1, it is natural to substitute -1 to get $f(-1)=2(-1)x^2+4(-1)-3$. Using the order of operations to simplify the right side, we arrive at f(-1)=-5. If you have no idea what is going on, here is an explanation. The AOS serves as the x-coordinate of the vertex. Since we know the x-value, we can find the y-value (***f(x) is the same as using the variable y because of ‘function notation’***). By plugging in the known x-value, we can find the related y-value and we now have an ordered pair. For this example the vertex is the ordered pair is (-1,-5). If you are trying to graph this problem, you should plot the point (-1,-5)

Lastly, we need to find two more points. To do this, pick any x-value to the left or right of the AOS. For this example I will use x = -3. Using the given function, $f(x)=2x^2+4x-3$, substituting -3 in for x gives $f(-3)=2(-3)^2+4(-3)-3$ applying the order of operations f(x)=3 and the ordered pair (-3,3). Again, because parabolas are symmetric, (-3,3) has a mirror image at (1,3). If you are graphing along with the post, plot the points and draw a curve to fit those points. Even though this is a sketch, you should try to make it neat.

Graphing a Parabola in Standard Form

## Using Properties of Parabolas to Graph a Parabola

When graphing a parabola it is good to know that it is symmetric. A parabola will have a vertex. The vertex is the minimum or maximum value of the parabola. It is good to note, the range has a restriction on it, not the domain. The vertex represents the axis of symmetry as I discussed in a post about modeling data with quadratic functions.

Starting with the standard form of a parabola $f(x)=ax^2+bx+c$, you can determine the where the vertex will lie in the coordinate plane and other aspects of a parabola. Below are the properties of a parabola or a quadratic function in standard form.

Given that $f(x)=ax^2+bx+c$ is a parabola when $a\neq0.$

1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
2. The axis of symmetry is given by the line $-\frac{b}{2a}$.
3. The vertex is given by the ordered pair $(-\frac{b}{2a},f(-\frac{b}{2a}))$
4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

The above information makes sketching the graph and finding the actual value of the maximum and minimum value of a parabola very easy. Just as you can write the equation of a parabola from three points, it is important to graph at least three points for a parabola.

Graph $-\frac{1}{2}x^2+2$. To graph any quadratic it is important to get an idea of what it should look like from the properties. In this case a < 0. Which means the parabola, will open downward. Often, parabolas are referred to as u-shaped graphs, so the u will open downwards. The axis of symmetry (AOS) will be at

$-\frac{b}{2a}=-\frac{0}{2*-\frac{1}{2}}=0$.

This function does not have a linear term, therefore the AOS is given by x = 0. Any parabola without a linear term will have an AOS of x = 0. Finding the vertex requires you to substitute 0 in for the variable x. This is where function notation comes in handy,

$f(0)=-\frac{1}{2}0^2+2$

$f(0)=2$.

Thus, the vertex $(-\frac{b}{2a},f(-\frac{b}{2a}))=(0,2)$

Because parabolas are symmetric, you need to only find one point other than the vertex. The third point is found by finding the corresponding point of the second point found. In this case, you would want to find f(2), which means when x = 2. The related y-value would be found with the following steps.

$f(2)=-\frac{1}{2}(2)^2+2$  substitute 2 in for x

$f(2)=-\frac{1}{2}(4)+2$  Square the 2

$f(2)=-2+2$  Multiply

$f(2)=0$  Add

Figure 1 Parabola y = (-1/2)x^2+2

﻿This gives a point of (2,0). The point the is symmetric about the AOS is (-2,0).All of the key points and the AOS are mark in Image 1 below. It is good to be worried about neatness when creating a graph, but if you are doing it by hand a good sketch will do. To the left is the best I could do using photo shop. If you need a an accurate graph, then using  graphing calculator or computer software would be your best bet.