What does linearize mean?

Basically when you linearize a data set you:
1. Graph log(y) vs. log(x), y vs. log(x) and log(y) vs. x.
2. If one of the graphs comes out linear, find the slope and vertical intercept and write down the corresponding linear function (e.g, log(y) = m log(x) + b, if the log(y) vs. log(x) graph is the one that comes out linear).
3. Solve the equation you get for y.
Here is a typical linearization problem, which should explain the process in some detail.

For the function y=f(t)=.93^t construct a table of y vs. t for t running from t=1.7 to t=2.9 in four equal increments. Using appropriate transformation(s) on the y column, the t column, or both, linearize this data set and demonstrate that the data set has in fact been linearized.

From t = 1.7 to t = 2.9 the change in t is 1.2. If the change occurs in four equal increments then each increment is 1.2 / 4 = .3.
So the t values are 1.7, 2.0, 2.3, 2.6 and 2.9. We get the following table:
t ** y=.93^t 1.7 0.883936388 2.0 0.8649 2.3 0.846273578 2.6 0.828048293 2.9 0.810215507.
Taking log(t) and log(y) we get
t * y=.93^t log t log y 1.7 0.8839 0.2304 -0.05357 2.0 0.8649 0.3010 -0.06303 2.3 0.8462 0.3617 -0.07248 2.6 0.8280 0.4149 -0.08194 2.9 0.8102 0.4623 ** -0.09140
We need to graph log(y) vs. log(t), log(y) vs. t and y vs. log(t) to see which, if any, of the graphs give us a straight line.
A graph of log(y) vs. log(t) is not linear--the points don't lie on or close to a straight line.
A graph of log(y) vs. t is linear--the points do lie on or close to a straight line. This line has slope -.0315 and vertical intercept 0, so we get the equation
log(y) = -.0315 t + 0 or just log(y) = -.0315 t. This gives us 10^(log(y)) = 10^(-.0315 t) or y = (10^-.0315) ^ t or y = .93^t.

How do you express the function y=f(x)=9*.95^x in the form y=A*2^kx?

9 * .95^x = A * 2^(kx) provided
A = 9 and 2^(kx) = .95^x.
The second equation is solved by taking logs of both sides:
log(2^(kx)) = log(.95^x) so kx log(2) = x log(.95). Dividing both sides by x log(x) we get k = log(.95) / log(2) = -0.074001.
So the y = A * 2^(kx) form is
y = 9 * 2^(-.074001 x).

Here is a similar question for further reference:

Express the function y=8*2^(1.03x) in the form y=A b^x, and also into the form y=A e^(kx). Y=Ab^x, y=8*2^(1.03x), would it not already be in this form? And I have no clue as how to put it in y=Ae^(kx) form?

y=82^(1.03x) = 8 (2^1.03)^x. Since 2^1.03 = 2.042, approx. the y = A b^x form is therefore
y = 8 * 2.042^x.
If 8 * 2^(1.03 x) = A e^(kx) then we have A = 8 and 2^(1.03 x) = e^(k x). Taking the natural log of both sides we get 1.03 x ln(2) = k x. Dividing both sides by x we get k = 1.03 ln(2), which we evaluate on a calculator to get k = .714 approx..
So the y = A e^(kx) form is y = 8 e^(.714 x).

When it says the graph depicts below...and then there isn't any graph shown, how would I answer that?

The best idea is to make up a reasonable graph and answer the question based on your graph. If you do well you get credit, if not the problem is omitted.

It says sketch a graph of the polynomial - 1.25(x-5)(x+2)(x-7), clearly indicating all intercepts and the large-[x] behavior of the graph.

At the 23-24-242 access site, check out the following links, which explain this process in several ways and with several examples:
12-01-2005_____questions_related_to_polynomials 12-03-2005_____zeros_of_a_polynomial___forms_of_a_polynomial 12-06-2005_____graph_of_a_polynomial_with_distinct_linear_factors 12-06-2005_____how_to_find_zeros_of_linear_polynomials 12-06-2005_____what_do_you_mean_by_a_quadratic_polynomial 12-06-2005_____explaining_the_graph_of_a_polynomial_function 12-06-2005_____quad_polynomial_with_no_zeros_is_not_product_of_two_linear_functions 12-06-2005_____what_quadratic_function_describes_behavior_of_polynomial_near_degree_2_zero

I understand that a graph needs to be made but I don't understand exactly how to go about doing it?

What does it mean by what specific function y=Ab^x+c fits the data points (4,1.635)and(9,.7663),with asymptote y=0?

Check out the link
12-06-2005_____what_exponential_function_with_asymptote_0_fits_two_given_data_points
at the access site for a detailed example.

For y=f(x)=3*.93^x find the ratio f(x+1)/f(X).

What does it mean by ratio?

The ratio of quantity A to quantity B is A / B.
For this function
f(x+1) / f(x) = 3 * .93^(x+1) / ( 3 * .93^x). 3 / 3 is 1 so this is equal to .93^(x+1) / .93^x. By the laws of exponents .93^(x+1) = .93^x * .93^1, or just .93^x * .93, so we have .93^x * .93 / (.93^x), which is just equal to .93.

If we do badly on this test can we retake it next week?

You can retake any test up through the end of the week. I'll count your best attempt.

What is a 75 in your grade book?

70-80 is the range for C.

What are the basic points of the exponential function y=f(x)=4*1.19^x? Graph the function using these points.

Are the basic points the same as the basic exponential function?

The basic points of an exponential function are at x = 0 and x = 1, and also the horizontal asymptote.
For y = 4 * 1.19^x the basic points are (0, 4), (1, 4.76) and the negative x axis as asymptote.

Another question related to your is the following:

What are the basic points of the exponential function y=f(x) =4*2^x? Graph the function using these points.

I also have no idea what to do on this one.

This exponential function has horizontal asymptote y = 0 --i.e., the x axis, which it approaches for negative values of x.
Its basic points are the x = 0 and x = 1 points.
Plot these points and draw an exponential graph through them with asymptote along the negative x axis.

The following links all address the basic points picture:

12-09-2005_____basic_points_of_exponential_function 12-06-2005_____graphing_exponential_function_using_basic_points 09-28-2005_____clarification_on_basic_points 09-26-2005_____brief_outline_of_scheme_of_the_four_basic_fn_families 09-21-2005_____plotting_four_basic_power_functions 10-05-2005_____basic_points 10-04-2005_____what_are_the_basic_points_of_a_linear_function 09-21-2005_____graph_3_basic_points_then_sketch_graph_based_on_those_points

What does this mean exactly: If Q(t)=Q0*.966^t, estimate to 2 significant figures the time necessary for the original quantity to dwindle to .04 of its original value?

The original value, at t = 0, is Q(0) = Q0 * .966^0 = Q0.
.04 of the original value is therefore .04 * Q0.
Q(t) = .04 Q0 when Q0 * .966^t = .04 Q0. We solve this equation for t. First divide both sides by Q0: .966^t = .04. Then either logs to find the value of t: log(.966^t) = log(.04) so t log(.966) = log(.04) and t = log(.04) / log(.966) = etc. (evaluate this on your calculator and you'll have the answer).
See also the link
12-06-2005_____estimate_time_for_exp_fn_to_reach__01_Q0

For the function y=f(t)=.93^t construct a table of y vs. t for t running from t=1.7 to t=2.9 in four equal increments. Using appropriate transformation(s) on the y column, the t column, or both, linearize this data set and demonstrate that the data set has in fact been linearized.

9 * .95^x = A * 2^(kx) provided A = 9 and 2^(kx) = .95^x. The second equation is solved by taking logs of both sides: log(2^(kx)) = log(.95^x) so kx log(2) = x log(.95). Dividing both sides by x log(x) we get k = log(.95) / log(2) = -0.074001. So the y = A * 2^(kx) form is y = 9 * 2^(-.074001 x).

Here is a similar question for further reference:

The best idea is to make up a reasonable graph and answer the question based on your graph. If you do well you get credit, if not the problem is omitted.

Check out the link 12-06-2005_____what_exponential_function_with_asymptote_0_fits_two_given_data_points at the access site for a detailed example.

The ratio of quantity A to quantity B is A / B. For this function f(x+1) / f(x) = 3 * .93^(x+1) / ( 3 * .93^x). 3 / 3 is 1 so this is equal to .93^(x+1) / .93^x. By the laws of exponents .93^(x+1) = .93^x * .93^1, or just .93^x * .93, so we have .93^x * .93 / (.93^x), which is just equal to .93.