Section+3.3

math {}$ Today we discussed $\S $3.3. Ms. Tyson started out the class with having all the groups work through $\S$ 3.3 In class experiment #1. This problem addressed finding the average rate of change. A method to solving the problem was discussed below the actual problem and walked through the steps if a student was having trouble figuring out a way to do it.

Then the class discussed the de{}finition of the average rate of change of $f(x)$ which is \begin{itemize} \item\emph{ Let f be a function and suppose A and B are distinct points on the graph of $y=f(x)$. The line passing through A and B is called a scecant to the graph of $y=f(x)$. Its slope is the average rate of change of $f(x)$ with respect to x between A and B.} \end{itemize} Also remember that the equation for the slope of a line is, $ slope=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$.

Ms. Tyson then reminded us that the word secant is a word we are familiar with. We have previously seen it in geometry where it means a line that intersects a circle twice, but here instead of a circle, it is intersecting a curve twice.

We then did $\S$ CYU #1. 1a. The graph of the function $f(x)=x^{2}$ with points A and B looks like this: $ math



math {}$The slope of the line, as you can see, is 7. That is also the average rate of change between the points A and B. 1b. To the find the equation of the secant line through $(a, f(a))$ and $(b, f(b))$, let's first look at the graph. $ math



math {}$ We used point slope form, $ (y-y_{1})=m(x-x_{1}})$ to figure out the equation for the line. Picking one point (we used $f(a)=-3)$, one first finds the coordinates. For $f(a)=-3$, the coordinates are $(-3, f(-3))$ or $(-3, 9)$. Then you plug those values into the point slope equation, so, $ (y-9)=m(x-(-3))$ which is the same as $(y-9)=m(x+3)$ Then we know that $$(y-9)=8(x+3)$$ $$y-9=8x+24$$ $$y=8x+33$$ which is the equation for the secant line. 1c. is done the same way as another example we did in class. For this example, it is important to keep a couple ideas from last year in mind. When you divide $f(x)$ by $(x-a)$ the remainder is $f(a)$.

$f(x)= Quotient(x-a)+remainder$

$f(a)=Quotient(0)+remainder$.

The example for 5th hour was: Let $f(x)=x^{3}+x^{2}-2$ Solve for the secant function. Divide $f(x)$ by $(x-a)$ (or $(x-b)$). So for this problem, divide $$f(x)$ by $(x-(-1))=(x+1)$$

Using long division, one arrives at the answer, $x^{2}$ with remainder 2. This means that $$f(x)=x^{3}+x^{2}-2$$ can be written as $$f(x)=(x+1)(x^{2}-\frac{2}{x-1})$$ and when you distribute the $(x-1)$, it looks like $$f(x)=((x+1)(x^{2})-2)$$ Then you plug in a value for $x$. Using $f(-1)$, one gets: $$(f-1)=(-1+1)(x^{2})-2$$ Which is the same as: $$f(x)-f(-1)=(x+1)x^{2}$$

$$\frac{f(x)-f(-1)}{(x+1)}=x^{2}$$ And that is the secant function for that equation!

\begin{itemize} \item {\bf Mrs. Tyson:} I took some of the strings of math and converted them to displayed equations so that the breaks between the lines are easier to see. You display the math by using 2 \$'s instead of just one around the mathematics. You might want to see if there is more math that might benefit from this. \end{itemize} $ math