Algebra: functions, Cartesian coordinates, lines and graphs

Algebra: Functions.

While I've mentioned up until now polynomials which are expressed in terms of a single variable and solving equations where polynomials in the form ax² + bx + c are set equal to zero and solving these for zeros of the polynomial or solutions to this equation, I haven't mentioned more generally the functional nature of the such equation.

A polynomial of this form or bx + c can be also called a function.  Sometimes functions will be written either as an equation of the form of two variables.

Examples

y = bx + c
y = ax² + bx + c

or they can be expressed in the written form

f(x) = bx + c
f(x) = ax² + bx + c

In this case we you might also see notation for the function f where
f:x  → y   This is to say the function f maps the values of x to y. 

As in the case above, where we directly write y = bx + c or y = ax² + bx + c similarly we are expressing the relationship between the variable y and x. Interchangeably it is much the same for the functions f(x) which is the resultant image map of x value mapped to the range of y. Thus f(x) can be thought the same with a y position, although the 'domain of x' may not be the same as the range of y, by this we mean that the set of points that are mapped from x may not equal the set of points mapped to, or in other words, while you might be able to arbitrarily chose positions of x expecting to find the results of mapping this same domain of x from the range of y may be another matter.

Example:

Consider the simple function

y = x²

If I am restricted to real numbers in the range of y, and I choose the value – 1, I could see that

-1 = x²

an additional property concerning exponents will be mentioned here. 

Namely, (x^y)^z =  x^y*z  
or
 (x^y)^1/z = x^y/z

but applying the multiplicative  rule for exponents to solve for x means taking the root of both sides of the equation (use the multiplicative rule for exponents in this case, we need x² in the form x¹, so (x²)^1/2 = (-1)^1/2 means x = plus / minus (-1)^1/2

Note that in the case to reduce a variable to single order variable in this case, taking a particular root to the nth order if you have such variable expressed as a single term on one side of your equation. We can conveniently do this more easy since we have no other lower order terms which would complicate the procedure of solving for x here in this more special case. For the root of a second order equation, just as in solutions for the quadratic equation there will be two solutions provided as given by the degree or order of the term depending on the whether the nth root is a even or odd integer. So in the case of 2nd degree term we expect a positive and negative solution. The reason for this relates to the mirror symmetry of the function itself owing to the even nature of the 2nd root: for example, x = 1 maps to the same y value which is 1² = 1 as does x = -1 which is (-1)² = 1 (remember negative times negative numbers invert the sign of the resultant product).

Therefore two solutions for the value of y = 1 apply in so far as values of x which is 1 and -1. As it turns out quadratic functions have this relation. A cubic doesn't do this so y = x³ has independent values of y for the x = 1 and x = -1. (remember -1*(-1)*(-1) = -1).

Getting back to our original problem where y = -1 and y = x², however, we notice we are taking the square root of -1. What is this? As it turn out the nth even root of any negative number is called an imaginary number in mathematics implying that the domain of x must require such that this number squared is a negative number, but no positive or negative real squared equals a negative number in the set of real numbers.  The function f: x  → y given by f(x) = x² maps all reals from x into the positive reals of y for the range, but the inverse function of f as it turns out maps can only map the positive reals back into the domain x.

While it would be getting into more advanced topics determining the what the range of y should look like under the image of mappings from the domain x to y for any given function expressed as a polynomial. I would say that at least when gaining some sense of a function a proper order at the level of elementary algebra is applied in so far as picking values, or at least incidentally through some guess of our own, we found out that y = -1 in the above case, doesn't have a real x mapping which leads to such value. And we have found out something of the range of our function inadvertently. Typically in high school algebra, where you knew that the domain were open in so far as values, we may be more likely able to pick arbitrarily values of x and solve for positions of y, and generally find the zeros of the function (polynomial) and solve for corresponding values of x. Solving for zeros may be the more rare exception where we pick a value for y or f(x) and the corresponding position x, as opposed to arbitrarily picking any value of y and solving x for the sake of elementary explorations in algebra. When dealing with single ordered equations the only exception to this is where f(x) = c meaning all values of x map to a constant value, and no other value of y would have a real x to map to such value c. Otherwise, one could likely substitute arbitrarily any position of x or y to find the other in the cases of polynomials of the form bx + c.

In elementary algebra, the Cartesian coordinate system which describes positions given by x and y values found in a coordinate pair (x, y) . If you have seen graph paper, then given a line moving in one direction on the grid defines the x axis, this corresponds to the horizon, and another line moving orthogonal (or perpendicular) to such line represents the y axis. Aligning your paper right, one such line would move away from you and into you if it were set on a table while the other would move parallel in front and to the right and left of you.

Sometimes also one can think of this conveniently as x axis being right to left, and y axis up and down, if this provides some perspective about the axis. Positive numbers would be expressed from an origin 0 to the right of the zero point on such axis, while negative numbers expressed to the left of such origin 0. While on positive y values would be expressed, above the origin 0 and negative numbers expressed below the origin 0. One procedure of plotting the position is given by counting the number of position to right or left as necessary to match the value x, and then plotting in the corresponding direction the value of y given the signs of each x and y value.

Thus if you had a 1 to 1 scaling or 1 tick for every integer position to the left or right of origin 0, and above and below such origin 0, the position (4,3) would plotted 4 ticks to the right of zero to the position (4,0), and 3 ticks up from 0 on the y axis above the point (4,0) which is (4,3).

For functions while we could pick many points and calculate the positions y, and then write out all the corresponding coordinate pairs so that we could plot them. In elementary algebra, the polynomial bx + c it turns out is a predictable graph in so far as its function. As it turns out this is a line, and this means that a line doesn't have so much unpredictable curvature that would require us to plot out any number of points to see what its graphical behavior might look like. If you've heard the term slope of a line, then this would indicate part of such behavior of the function, and the slope can simply be computed by plotting two points or in examining the given polynomial written in its simplified form bx + c as shall be explained later.  All other points on the line will have the same slope when the line runs through them. To calculate the slope we simply compute the difference (or deltas) of y and x values, and compute the ratio of both. The Slope of a line is given by m = (y₂ – y₁) / (x₂ – x₁) is equivalent to delta y / delta x

Where points (x₁, y₁) and (x₂, y₂) are such that x₁ is not equal to x₂ and y₁ is not equal y₂, or in other words these are distinct points.

There are perhaps two important aspects of the graphical to equation and equation to graphical approach when dealing with functions and their graphs.

If we know the graph and we need the equation or the function (of the line) which describes the points on the graph, then we could do so, in the following manner:

1. Record to distinct points on such line, recording these values as (x₁, y₁) and (x₂, y₂) respectively. Then compute the slope m by the equation shown above.
2. Find the y-intercept b of the line. This is the position (0, y) on the line, or in other words, this is the position where the line intercepts the y axis at the x position equal to zero. Again we'll label the y intercept b for our purposes.
   3. The equation for the line is given by y = mx + b. So solve by substituting the values of m (computed in step 1) and b (computed in step 2) into the equation y = mx + b.

This method allows you to find the equation for the line, if only the graph of such line is shown.

If you have the equation but need want to show line graphically, then it suffices to do the following:

1. Pick arbitrarily one value x. From the equation in the form mx + b compute the value y. Plot this point
2. The slope of the line is given by m in the equation y = mx + b, so we have this computed for us. Positive slopes incline from decreasing to increasing y in the positive direction of x (moving to the right of a given position position plotted in 1), while

negative slopes increase in the negative direction of x axis moving to the left of the plotted position in step 1. If m is a fraction, and the fraction is simplified meaning that no common divisor exists between the numerator and denominator ( or at least I recommend this), one can plot the rise (delta y) over run (delta x) by the rise given by the integer in the numerator, followed by the run given by the integer found in the denominator. Thus a slope of m = 4/5 would have a rise of 4 ticks up from the position y1 from the plot position from step 1 (which is a position (x₁, y₁+4)), and then one would count 5 ticks to the right of this position which ends at (x₁ + 5, y₁ + 4). Then draw a straight line through both such positions. This line should continue also so that the when x = 0, the intercept position b should be given for the y position.

Keep in mind if the slope m is a whole number you'd this would need be converted to a fraction to find the slope of a line.

Example, m = 3 is equivalent to m = 3/1. The rise is 3 (delta y) , the run is 1 (delta x).

If we are plotting from a known point using such slope, we count 3 ticks up from its position which ends at position (x₁, y₁ +3), then we count to the right of such position 1 tick ending at position (x₁ +1 , y₁ +3).

Example, m = -4 is equivalent to m = -4 /1. The rise is -4 (delta y) , the run is 1 (delta x). If we are plotting from a known point using such slope, we count -4 ticks down from its position which ends at position (x₁, y₁ – 4), then we count to the right of such position 1 tick ending at position (x₁ +1 , y₁ -4).

It helps to have a ruler, pencil, and graph paper handy when plotting functions here.

In a last bit, it could be mentioned that formulation for the slope of a line, is similar in the analysis approach to finding the tangent of a line at a point by way of derivatives, except technically in calculus one defines the line between two such points (x₁,y₁) and (x₂, y₂) secant.  In calculus if the function is determined continuous and behavior of a function appears otherwise orderly enough as one were to approach a point calculating a the slope with a given neighborhood point more closely in successive steps, one would find the secant lines between point and neighborhood points approaching the tangent or slope at such point, or otherwise called the derivative value of the function at such point.   While the tanget or slope and secants in the line bx + c is fixed the same for all such points, the function f(x) = x² varies from point to point...see for yourself if you like (find the secant line of any given (x₁, y₁) and (x₂, y₂) using the slope formula defined above and try another two sample points, see if the secant values match up).  Functions with varied tangent (slope) curves implicitly provides just some of many motivations surrounding Calculus as a branch in mathematics.