How do we find an equation for the straight line through two known points?

Use the slope = slope form, or substitute your coordinates into the form y = m x + b and solve for m and b.

In some problems you are asked to give the details of the slope = slope form, including a graph.

The graph (which is in your worksheets) consists of the points (x1, y1), (x2, y2) and (x, y), all depicted on a straight line. The straight line has constant slope, so (x, y) will be on the same line as (x1, y1) and (x2, y2) if, and only if, the slope from either of these points to (x, y) is equal to the slope between the two points.

So, for example, (x, y) is on the line if and only if (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). This is the slope = slope form.

Suppose you want to find the equation of a line through (x1, y2) = (2, 3) and (x2, y2) = (5, 7). The slope = slope equation would be (y - 3) / (x - 2) = (7 - 3) / (5 - 2). You would first express the equation in this form, including a graph. Then you would simplify and solve for y.

You could also find the equation by substituting the coordinates of the points into y = m x + b. You would get

3 = m * 2 + b and 7 = m * 5 + b.

You would solve the equations simultaneously for m and b then substitute back into the form y = m x + b.

The resulting equation will be exactly the same as the equation you get when you solve the slope = slope equation (y - 3) / (x - 2) = (7 - 3) / (5 - 2) for y.

For n = 1 we get a(1) = a(0) - 2 * 1^2 = -2 - 2 = -4 For n = 2 we get a(2) = a(1) - 2 * 2^2 = -4 - 4 = -8 For n = 3 we get a(3) = a(2) - 2 * 3^2 = -8 - 18 = -26 etc.

For more detail on recurrence relations see also the following links at the 23-24-242 access page:

10-27-2005_____recurrence_relation 10-29-2005_____recurrence_relation 11-02-2005_____recurrence_relation 11-08-2005_____ 11-04-2005_____getting_the_270th_member_of_a_sequence_defined_by_a_recurrence_relation

In the first place we have to remember that x^-p = 1 / x^p.

For any power function we use the basic points at x = -1, 0, 1/2, 1 and 2.

For a negative power function the x = 0 point is undefined, since we can't calculate 1 / 0^p = 1 / 0. (The reason: You can't count to 1 by zeros, so there is no answer to the question of how many times 0 goes into 1).

For values of x very near 0 the magnitude of 1 / x is very large, and the close x is to zero the larger the magnitude. (The reason: If you count to 1 by small increments, it takes a lot of them. If you want an example, count to 1 by .0001. You start .0001, .0002, .0003, ... and it takes a good while to get to 1. So 1 a small number goes into 1 a lot of times).

So the vertical line x = 0 is a vertical asymptote for the graph.

For integer powers the graph will either be symmetric with respect to the y axis or antisymmetric with respect to the y axis. Comparison of the basic point at x = -1 with the basic point at x = 1 tells you which.

Also for large magnitudes of x, x^p will be large so 1 / x^p will be close to 0, since 1 divided by a large number is close to 0. This means that the positive and negative x axes will be horizontal asymptotes for a negative-power function.

Examples:

-1^-2 = 1 / (-1)^2 = 1 / 1 = 1. (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 1 * 4/1 = 4. 1^-2 = 1 / 1^2 = 1 / 1 = 1. 2^-2 = 1 / (2^2) = 1/4. So

For the function y = x^-2 the basic points are (-1, 1), (1/2, 4), (1, 1) and (2, 1/4).

There is a vertical asymptote at x = 0--i.e., at the y axis.

The x = -1 and x = 1 points have the same y value so the function is symmetric with respect to the y axis.

Plot the points and use common sense.

-1^-3 = 1 / (-1)^3 = 1 / 1 = 1. (1/2)^(-3) = 1 / (1/2)^3 = 1 / (1/8) = 1 * 8/1 = 8. 1^-3 = 1 / 1^3 = 1 / 1 = 1. 2^-3 = 1 / (2^3) = 1/8. So

For the function y = x^-3 the basic points are (-1, -1), (1/2, 4), (1, 1) and (2, 1/4).

There is a vertical asymptote at x = 0--i.e., at the y axis.

The x = -1 and x = 1 points have opposite y values so the function is symmetric with respect to the y axis.

Again plot the points and use common sense.

Better give me a specific example of this. I can't tell from your question what you are asking about.

y vs. x means y on the vertical and x on the horizontal axis. x vs. y would put x on the vertical and y on the horizontal, which is not the traditional way of graphing functions. The quantity before the 'vs.' is on the vertical.

Generally y will stand for the independent variable and x for the dependent variable. It's usually but not always easy to tell which is independent and which dependent. For example depth depends on the clock time; the clock keeps going independent of what happens in the cylinder. The period of a pendulum depends on its length, not vice versa.

Applying this convention to the situation of your question:

If the graph is of depth vs. clock time, then depth is the y coordinate and clock time t is the x coordinate. Depth goes on the vertical axis and clock time on the horizontal.

y = log(x) means 10^y = x, and vice versa.

So 100 = 10^2 is of the form x = 10^y for x = 100 and y = 2, which by the above means that y = log(x), or 2 = log(100).

We can find the exact value of log{base 8}(1024) since 8 and 1024 are both powers of the same number, namely 2. This is so because 8 = 2^3 and 1024 = 2^10.

log{base b}(x) = y

means exactly the same thing as

b^y = x.

It follows that log{base 8} (1024) = y if

8^y = 1024.

Expressing 8 and 1024 as powers of 2 we have

(2^3)^y = 2^10. Using the laws of exponents this becomes 2^(3y) = 2^10. We conclude that 3y = 10 so that y = 10/3.

An irreducible quadratic factor is a quadratic expression that cannot be equal to 0. You determine if a quadratic is reducible or irreducible by setting it equal to 0 and solving the resulting equation, usually by means of the quadratic formula.

If you get one or two real solutions z1 and z2 then the expression is reducible, and has factors (x - z1) and (x - z2).

If you get imaginary or complex-valued solutions then the expression is irreducible.

For example x^2 + 3 x + 2 is reducible since it can easily be factored into (x+1)(x+2). The solutions of x^2 + 3 x + 2 are -1 and -2, which can be found either by the above factoring or by the quadratic formula.

x^2 + 3 x + 1 is also reducible, but you'll never factor it by trial and error. The reason is easy to see if we solve the equation x^2 + 3x + 1 = 0 when x = (-3 +- sqrt(3^2 - 4 * 1 * 1) / (2 * 1) = -3/2 +- sqrt(5) / 2. The solutions are z1 = -3/2 + sqrt(5) / 2 and z2 = -3/2 - sqrt(5) / 2, and the factors are (x - (-3/2 + sqrt(5) / 2) ) and (x - (-3/2 - sqrt(5) / 2) ).

x^2 + 2 x + 3 is not reducible, since the solutions to the equation x^2 + 3 x + 2 = 0 are x = (-2 +- sqrt(2^2 - 4 * 1 * 2) / (2 * 1) = (-2 +- sqrt(-4) / 2 = -1 +- i, where i = sqrt(-1). These are not real-number solutions; we don't get a real number since sqrt(-4) is not a real number. Thus x^2 + 2 x + 3 is an irreducible quadratic.

The fundamental theorem of algebra says that any polynomial can be factored into linear factors of form (x - z) and irreducible quadratic factors of form (a x^2 + b x + c) in which b^2 - 4 a c is negative.

The key to graphing polynomials is that the linear factors give us zeros, while the irreducible quadratic factors do not.

The following links might also be relevant to your question:

12-01-2005_____questions_related_to_polynomials 12-06-2005_____graph_of_a_polynomial_with_distinct_linear_factors 12-03-2005_____zeros_of_a_polynomial___forms_of_a_polynomial 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 12-09-2005_____graphing_a_factored_polynomial_with_negative_leading_coefficient 12-09-2005_____graphing_a_specific_factored_polynomial 12-09-2005_____meaning_of_distinct_and_repeated_linear_factors_and_their_effect_on_a_graph