And on those worksheets question 5, I'm a little confused about finding y=f(x) when we're supposed to be substituting in for example for how long was the depth between 34 and 47 centimeters? here is what i did - y=f(47-34)

by how much did the depth change between t=23 sec & t=34sec 34-23=f(x)

Is this correct?

This problem is a little complicated and I don't really expect a great answer at this stage. But it is y that is changing from 34 cm to 47 cm, so you have do to this:

Solve the equation 34 = f(t) for t.
Solve the equation 47 = f(t) for t.
Then find the difference between the two values of t.

More specifically you would substitute the first y value in the specific quadratic you got for your depth vs. clock time model and you would solve the equation for t. Then you would do the same for your second y value, and find the difference in t values.

And on number 6, what do you mean when you say determine the average deviation for your model and graph your function?

I picked my three points but I'm unsure of what to do from there

Go ahead and solve the 3 equations you get for a, b and c to get your model.

Then plug each of the t values into your model and see what you get for your y values.

Finally plug in each of the t values in the data set to get your model's predictions of the y values, and see how close each predicted value comes to the data.

The average magnitude of the difference between your model's y values and the observed y values will be the average deviation.

See also the worksheet 'Completion of the Initial Flow Model' under the heading 'Assessing the Function Model'.

Will there be a quiz over the information contained in the final portion the function notation packet despite the fact that the exercises were not assigned for friday's class?

I won’t grade you over any questions about material that hasn’t been covered or assigned.

If I do ask you such a question it won’t be counted as part of your quiz grade. For example I might well to ask you tomorrow to find the slope of a graph, in order to see how many people can answer it, but if it isn’t directly related to previous assignments it wouldn’t count toward a quiz grade.

My question might seem a little out of place, but it is a question that no math teacher that I have ever had has sufficiently answered for me, and it is something that plagues me every time I do homework for this class. Why is it that any number raised to the power of zero, is one? This doesn't make sense to me and my previous teachers simply told me to memorize it as a rule. I don't understand how something raised to the power of nothing equals one. Any enlightenment that you can shed on this would be greatly appreciated.

This is a great question and I'm glad to answer.

You will accept that x^2 * x^3 = x^5. This is as it must be, since x^2 = x * x and x^3 = x * x * x so that x^2 * x^3 = x^5.

You will also accept that x^2 * x^-5 = x^-3, since x^2 = x * x and x^-5 = 1 / x^5 = 1 / (x * x * x * x) so that x^2 * x^-5 =
x * x / ( x * x * x * x * x ) = 1 / (x * x * x) = 1 / x^3 = x^-3.

These example illustrate why x^a * x^b = x^(a + b).

Ok. So now what is x^2 * x ^ -2. You immediately see that we get x * x / ( x * x ) = 1. You also see that we get x^(2 - 2)
= x^0.

Thus x ^ 0 = 1.

This is true no matter what value x takes.

And it follows directly from the laws of exponents.

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I was just wondering about those worksheets you gave us. I attempted all way through all of the 10 exercises but I'm slightly confused on some of the questions..." "Like on number nine whcih states if f(x)=2^x then writes expressions for...
f(x)+3 would this one look like this? f(x)+3=2^(x)+3 with the three actually just being added to the final equation or would the three be added to 2^(x+3)

f(x) = 2^x so f(x) + 3 would be 2^x + 3. You just substitute 2^x directly for f(x) and don't read anything into the
expression.

By contrast f(x+3) = 2^(x+3). Here the x + 3 is substituted for x.

On number ten would I just use the formula for the generalized power function and plug those numbers in? Its asking the illumination y from a certain florescent bulb is given as a function of distance x by the generalized power function for p=-1 with A=370 (by the way, what does a capital a stand for?)

h=0 and c=0 Determine the illumination at distances of 1 2 3 4 units... would I just plug everything in and plug in the units as x?
fs
Capital A is generally used in this course to denote the vertical stretch.

The generalized power function is y = A ( x - h) ^ p + c. SO if A = 370, p = -1, h = 0 and c = 0 the function is

y = 370 ( x - 0 ) ^ (-1) + 0, or just
y = 370 x ^ (-1). This is the power-function form. It just means

y = 370 / x.

Illumination y is a function of distance x. So x stands for distant, and distances of 1, 2, 3 and 4 units stand for x = 1,
2, 3 and 4.

Thus you would plug each of these number, in turn, into the function to obtain the corresponding values of the illumination
y.