If a zero at x = z is repeated twice, that means that the linear factor (x - z) occurs twice in the factored form of the polynomial, giving us the factor (x - z)^2.

This causes the graph to behave much like a parabola A ( x - z )^2. The quadratic function y A ( x - z)^2 is horizontally shifted so its vertex is at (z, 0), so its graph just 'touches' the x axis at its vertex. So the graph of our polymonial just touches the x axis at its zero, in the manner of a parabola.

If the zero was repeated n times, then (x - z) would occur n times in the factorization of the polynomial and would therefore contain the factor (x - z)^n. The polynomial would therefore act in the vicinity of x = z like a power function A ( x - z)^n, passing through (z, 0) in the same manner as the power function y = x^n passes through the origin.

For radioactive decay the growth rate is negative, giving us a growth factor less than 1. So the function could be expressed as A(t) = A0 * b^t, where the growth factor b is a positive number less than 1.

The halflife is analogous to the doubling time. It is the time required for the function to reach half its value.

Just as the doubling time of an exponential is the same no matter at what time you start, the halflife is also consistent.

The function A(t) = A0 * b^t has halflife tHalf, where

A(t + tHalf) = 1/2 A(t), giving us A0 * b^(t + tHalf) = 1/2 A0 * b^(t). Since b^(t + tHalf) = b^t * b^tHalf, we have A0 * b^(t) * b^(tHalf) = 1/2 A0 * b^t. Dividing both sides by A0 * b^t we get b^(tHalf) = 1/2. Taking logs of both sides tHalf log(b) = log(1/2) so that tHalf = log(1/2) / log(b).

The function y = log{base b}(x) has base b. The function y = ln(x) has base e. The function y = log(x) has base 10.

This is something you would likely get after using logs.

10^b is just a number, which we might call A. So let A = 10^b.

10^(m t) could be written as (10^m)^t or as (10^t)^m. Assuming t is our variable, we use the first form and let 10^m = d (would have used 10^m = b but we've already got b in the original equation). Then 10^(m t) would be just d^t.

Now our equation is in the form

y = A * d^t,

which is the form of an exponential.

There are lots of ways to get a quadratic function. The depth vs. clock time model we did early in the course is one example.

No matter how the quadratic function comes about, its factored form is one of the following three:

The factored form could include two distinct linear functions. For example x^2 + 5 x + 6 = (x + 2)(x+3). In this case the function has two distinct zeros and its graph passes through these zeros.

The factored form could include the same linear function twice. For example x^2 + 6 x + 9 = (x + 3)(x+3) = (x+3)^2. In this case the function has one zero. On its graph this zero occurs at the vertex, where it just touches but does not pass thru the x axis.

The only other possibility is that the function does not factor into linear functions. In this case the quadratic is said to be irreducible. This is the case if and only if a x^2 + b x + c = 0 has no solution.

See also the link 12-10-2005_____what_are_irreducible_quadratic_factors___fund_thm_of_algebra___graphing_polynomials.

The log and the exponential are inverse functions.

To get a log table from an exponential table you just reverse the x and y columns.

To get a log graph from an exponential graph you just reverse the x and y coordinates.

So log{base b}(x) = y if and only if b^y = x.

-b / (2a) is the x coordinate of the vertex of the parabolic graph of y = a x^2 + b x + c.

The line y = -b / (2a) is the axis of symmetry of the parabola.

Mass occupies volume.

Volume grows in 3 independent directions when you scale a figure up.

So the proportionality y = k x^3 dictates this situation, where y is volume and x is diameter.

Plugging 409.6 g in for y and 8 cm for x you get

409.6 g = k * (8 cm)^3 or 409.6 g = k * 512 cm^3 so that k = 409.6 g / (512 cm^3) = 82 g / cm^3, approx.

Now your proportionality reads

y = 82 g / cm^3 * x^3. Or if you want to be clear on what the variables mean you could write

V = 82 g / cm^3 * d^3.

The vertical stretch factor can make the parabola taller and thinner or shorter and fatter.

You get the shape of the parabola by applying the vertical stretch factor to the basic y = x^2 parabola, getting y = a x^2 with its basic points (-1, a), (0, 0) and (1, a). These three points give you a very good indication of the shape of the parabola.

These points can then be vertically and/or horizontally shifted to any point of the plane, giving us our vertex and points 1 unit right and left of the vertex.

The other form is y = A * e^(kx).

Note that y = A * 2^(kx) can be written y = A * (2^k) ^ x, which is just y = A b^x with b = 2^k.

Similarly y = A * 2^(kx) can be written y = A * (e^k) ^ x, which is just y = A b^x with b = e^k.

If we have a table of y vs. x values for the f(x) function, we just square the y column to give us the table of values of the squared function.

If we have the graph of y = f(x) vs. x we could make a table and then square the y values, which squared values we would then graph against x.

If we have a rule that allows us to find f(x) for a given x:

we find at some x the value f(x) of the function. We square that value to get (f(x))^2. So we say that f^2 ( x ) = (f(x))^2.

When we have done this for every x value in the domain of the function we have the values of our f^2 function.

If we have a symbolic formula for y = f(x), we square that formula to get the formula for y = f^2(x).

The y = log(x) function is the inverse of the y = 10^x function. The y = ln(x) function is the inverse of the y = e^x function.

An asymptote is a straight line that is approached closer and closer by the graph of the function, but which the graph never actually reaches.

1/2 and 2 are both in there.

First you get your function model for a set of y vs. x data.

Then you plug your x values into the function to get the y values predicated by the model. Ideally these should agree with the y values of your data.

However this is not typically the case. In most data there are random errors or fluctuations, so the predicted y values don't exactly match the y values of your data.

The differences between the observed and predicted y values are deviations.

See also

09-21-2005_____determine_the_average_deviation_for_your_model 09-21-2005_____average_deviation

f is the symbol used to invoke the function in question.

For example

if f(x) = x^2, then A f(x-h) + c = A * (x - h)^2 + c if f(x) = e^x, then A f(x-h) + c = A * e^(x - h) + c etc.

The graph of y = A f(x-h) + c takes every point of the graph of y = f(x) and does the following:

vertically stretches the point by factor A, moving the point A times as far from the x axis horizontally shifts the point h units, moving it h units in the x direction. vertically shifts the point c units, moving it c units in the y direction.

For the basic functions it is easy to follow the effect of A f(x-h) + c on the basic points, which gives you a good indication of what the graph of the new function will look like.

For other functions you can also use a set of representative points of the f(x) function to construct the A f(x-h) + c function.

Take the square root. Of course if the number is negative you don't get a real-number result.

A = pi r^2, where r is the radius of the circle.

The circumference of a circle is C = 2 pi r, which is often confused with the area.

You can keep them straight by understanding that area is measured in square units, so it has to involve r^2 rather than just r.

Circumference on the other hand has units of distance, the same as the units of r.