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Tuesday, December 2, 2014

Exponential growth/decay

Hi, I’m going to explain exponential growth and exponential decay. First of all, we should start with exponential functions. Exponential functions are functions that change at a constant rate. Exponential functions are usually in form .

Meanwhile, A cannot be equal to zero and B has to be greater than 0. When A is greater than 0 and B is greater than 1, the function has exponential growth. When A is greater than 0 and B is less than 1, the function has exponential decay. In the equation above, A represents the initial value, B represents the growth rate and x represents the time.

Exponential growth/decay has many real world applications and is therefore very important in today’s society. It can help us determine population growth, how much money we get from interest and it can help determine the half-life of things such as bacteria. These all have important real world implications, which is why we should delve further into the matter.

Lets looks at some examples of exponential growth. Suppose Joe makes $100,000 a year as an investment banker. Every year, Joe’s salary will increase by 12 percent annually. How would you write this formula? And what will Joe’s salary be in 8 years?
Joe has an initial salary of $100,000, therefore, A would be 100,000. Meanwhile, Joe’s salary increases by 12 percent annually. So B would equal 1.12. In order to find B, you have to add  1 to the growth rate. So it would be 1+.12 and that would give us 1.12 for B. Since we want to know what Joe’s salary in 8 years will be, X=8. Thus, the formula for this equation would be Y= 100,000(1.12)^8 and his salary in 8 years would be  $247,596. We also know this is an example of exponential growth because B>1.

The following video is a cool illustration of exponential growth.






For exponential growth/decay functions, I generally use two tricks. First, If I know the growth rate is more than 1, the function is exponential growth. That also helps you know how the graph of the function will look. The above graph (above link, I couldn't paste the graph) is an example of exponential growth. If the growth factor is less than 1, I know it will automatically know it will be exponential decay. The graph of an exponential decay function will look very different from the graph of exponential growth. The graph below (the link below, the graph wouldn't show on the blog) represents exponential growth. The initial value of both graph are very different to help show the impact of the growth factor. In the exponential growth graph, the initial value is relatively low and over time the value goes up because there is growth. In contrast, the exponential decay graph has a very high initial value but because there is a decay rate the value decreases over time.
The other trick has to deal with the growth factor. If I’m already given a percent, It seems easier for me. If the percent is over 100 it would be similar to getting a b value over 1 and I would know there is exponential growth. Likewise, if the percent is 70 for example, I know there is a decay rate because If I covert 70 percent to a decimal, I would get .7.


One last example

After Mary bought a car, she realized the value of her car would steadily depreciate. She bought her car for $45,000 and her car will depreciate by 10 percent annually. How much would her car be worth in three years?


 In the given example, we are told that the car will depreciate by 10 percent annually and we have an initial value of 45,000. The tricky part of this equation is that they say it will depreciate by 10 percent annually. If we convert that to a decimal we get .1 the next step would be to subtract 1 by .1 which will give us .9. This is because after one year, the car will be worth 90% of what it was originally worth. Therefore the growth factor in this equation would be .9  which means its exponential decay. Therefore, the equation of this function would be f(x)= 45,000(.9)^3 and that would give Mary a value of $32,805 after three years.

Monday, December 1, 2014

Compound Interest

Functions

A function is a relationship where there is exactly one output per input. There can be multiple inputs for one output. When a function is graphed, it passes the vertical line test, meaning that if I drew a vertical line through the graph, it would only touch the graph once. Functions can be represented on a graph, table, formula, or in words. The standard function notation is output = f (input). Below is an example of a graph of a function.
 


This is an example of a function in a table. You plug the x, which is the input value, into the equation below in order to get y, which is the output value.
F(x)= x+3
X (input)
F(x) (output)
3
6
4
7
5
8
6
9
7
10
8
11


Functions are useful because they show the relationship between dependent and independent variables. An example of this is the relationship between temperature and time of day, time being the independent (x) variable, and temperature being the dependent (y) variable since temperature is related to the time of day.

Learning Linear Functions- Jean Marcel Correa



Blog 4: Domain and Range

Today we will be learning Domain and Range. 






Blog Post # 4 Completing the Square

Today we will be completing the square.
We complete the square in order to solve a quadratic equation. 
Recall that the formula for a quadratic expression is ax2 + bx +c2.
In this lesson, we will be completing the square when the leading coefficient, “a”, is equal to 1.
Recall that a perfect square trinomial can be factored as (a+b)2 = a2 +2ab +b2. 
For example, (x+4)2 = x2 + 8x + 16.
In many cases however, the expression doesn’t have a perfect square trinomial.
In x2 + 8x – 33 = 0, we complete the square in order to get a perfect square trinomial. 
We do this by dividing “b” by 2 and squaring the answer [(b/2)2]. 
Here that is (8/2 = 4)2; 42 = 16. 
Once we have completed this step, we should isolate the “c” value.
x2 + 8x – 33 +(33) = 0 + (33). 
Now we have x2 + 8x = 33. 
We want to complete the square by converting “x2 + 8x” into a perfect square trinomial. 
Again, to do this we take half of b and square it, [(b/2)2] = (8/2 = 4)2; 42 = 16. 
Now we add 16 to both sides of the equation.
x2 + 8x + 16 = 33 + 16. 
This then equals x2 + 8x + 16 = 49. 
Now we can factor the perfect square trinomial on the left hand side. 
From factoring, we get (x + 4)2 = 49. 
We now have a quadratic that we can solve by taking a square root.
√(x + 4)2 = √49. 
So, x+4 = 7 or x+4 = -7.  
We then solve both equations for x.
(x + 4) -4 = (7) -4 => x=3 and (x + 4) -4 = (-7) -4 => x= -11.

You can and should check your answers by plugging them back into x2 + 8x + 16 = 49.