Section+3.10

In class we discussed with Ms.Tyson a bit about the problems we discussed in class yesterday. She explained why we could not use factoring to solve for f(x) math \[f(x)=\frac{x^2+x+1}{x-1}= \frac{x(x+1+ \frac{1}{x})}{x(1-\frac{1}{x})}\]

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Though this statement is completely true, it causes the end behavior of the graph to be different of the polynomial to be different. For the first polynomial, the one that is not factored, the end behavior is (x+2). The factored polynomial the end behavior is (x-1). The correct answer is the polynomial with the EBA of (x+2). The reason you can not factor out the x is that the two values influence each other at some point, and by factoring it you are changing the two values thus changing how they influence each other which changes the answer. The only way to solve the equation is to do it the long way through division.


 * Mrs Tyson: The previous paragraph is a little difficult to parse. There are some errors in the structure of the English. Hopefully this will be improved shortly. In the meantime, the take away message is to find the EBA of a rational function you must do long division! No shortcuts. RATS!

After clearing up the issues from yesterday we moved on to discussing the new section. Once again we saw a very familiar equation from Algebra 2.

math A=P(1+\frac{r}{n})^{nt}

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Which in this equation A is the money in the bank after the principle, P is compounded 1+ the rate of interest as a decimal, over n, the number of times interest is compounded a year. then it is put to the power of n the numbers of times interest is compounded in a year and the number of years that is in the money the bank collecting interest.

For example, let's say a person has twenty dollars. They put that twenty dollars in an account that promises twenty percent interest compounded monthly. She leaves that money in her account for three years. How much money does she have at the end of those three years. First you set up the equation. Which should look like this.

math \[A=20(1+\frac{.2}{12})^{12\times3 }\]

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Then you should find that A= $36.26, but what if they instead of having is compounded monthly it was done quarterly.

math A=20(1+\frac{.2}{4})^{4\times3 }

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A should now be equal to $35.92. The pattern continues as you place values in the equation.

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Now if you were to find a tangent line to any number you can use the secant line equation to help "approximate the equation. So, let's say you wanted to find the tangent line of the graph of $ 4^x

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The way to figure out the tangent line is actually by using the equation to find the secant line, but limit it to one point. You can do this by making the difference between $f(x_{1}) and f(x_{2})$ very small, so maybe by 0.0001.

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The equation of the secant line is

math \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}

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So let's say you want the line tangent to the graph at x=1.The equation for tangent line in this case is \frac{f(4^1^.^0^0^0^1))-f(4^1)}{.0001} $. The graph with the line and the estimated tangent is below.



This is the graph of the approximation. Now here is the actual tangent line



As you can see the graphs are pretty similar, there some differences but otherwise they are close enough that using the secant line equation is close enough that it makes a good guess of the tangent line of the graph.