A zero is a factor times negative one. If you set a factor equal to zero and solve, you get your zero. The degree of a polynomial tells us how many zeros it has. It does not always tell us the factors because sometimes a zero repeats, thus only giving us one factor for two zeros.
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Part 1: The GIF below shows the secant line used to find the average rate of change between the fixed point (2,2) and another point on the curve f(x)=.5x2. Using desmos, recreate this GIF, and answer the following questions in a blog post. y=kx-1 1)How does the slope of your secant line (your average rate of change) change as your roaming point changes? How is this related to the location of your roaming point on the curve? As the value of k increases, the slope increases. as the x value of the location of the roaming point, slope also increases. When K increases, the x and y value of the roaming point increase. 2)When is the slope of your secant line closest to the rate of change at the point (2,2)? When k equals 1.5 3)Is your average rate of change ever 0? If so, where does this occur? No, as long as x is increasing (which it is), your rate of change will always be a positive number in this function. It has to mathematically. Part 2: 1)How does the slope of your secant line (your average rate of change) change as your roaming point changes? How is this related to the location of your roaming point on the curve? The answer here is the same as in part 1. The rule is the same, just with a negative reflection. As k increases, so does the slope. 3)Is your average rate of change ever 0? If so, where does this occur? No, the graph maintains either a positive or negative rate of change. http://goo.gl/tHQxP9 Blog post: Create a new post titled “Piecewise Functions”. Please insert a screenshot of graph #4. Describe in your post how you wrote a function to describe the behavior of the graph and how the domain and range changed from each individual function. When writing this function I had to consider domain. It was split up into (-infinity,0), (0,2), and (2,infinity). For the (-infinity,0) I knew it was linear so I started with that. I also knew it was negative and i saw where my Y intercept was. Observing the slope was the third part of the puzzle. For the (0,2) I noticed that it looked like a square root function and was. Seeing the Y intercept and slope (rate of change) in the function, it led me to educated guess and check. The same goes for (2,infinity) except I didn't have a Y intercept to work with. All I knew is that it would be negative. The domain was split up into the three functions, but the range was different. Starting from right to left, the ranges were (2,infinity),(0,2), and (0,infinity) Here is another piece wise function that I thought illustrated what it is well.
In this activity, I learned quite a bit about the math behind even and odd functions. It has more to it than just the highest power of x. Even and odd functions are similar in that proving them to be one or the other algebraically is almost the same. The only difference is a negative sign. Proving functions are negative is -f(x)=f(-x). The positive version is f(x)=f(-x). The functions themselves are similar because they have symmetry. The even functions are completely symmetrical while the odd functions are sort of symmetrical. Even functions are (x,y) and (-x,y) while negative functions are (x,y) and (-x,-y). I'm not sure it's correct to say they are "kind of" symmetrical, but the points on each side have a correlation with the points on the other side in an odd function. That's another thing both functions have in common; the points on one side have a correlation with the points on the other side.
Even Odd |