Module 2 Study Guide The second module covers the following sections of the textbook: 3.3-3.7, 4.1, 4.2, 4.5, and 5.1-5.3 Sections 3.3-3.6 This is a continuation of the study of linear functions that we started in the first module. Specifically, we're covering graphing linear functions by multiple methods (points, intercepts, slopes), working with slope, computing the equation of a line given some information about that line, and working with parallel and perpendicular lines. Graphing Linear Functions This is a continuation of objective B in section 3.2. Make sure that, given an equation, you can determine that it's LINEAR FUNCTION based on its characteristics. Namely, you should be able to rewrite it in y=mx + b, or f(x) = mx + b, form. For instance: 3x + 2y = 6 is a linear function because it can be re-written as y = 3 2 x + 3, but y=x 2 5 is not linear because of the "squared". One strategy for graphing linear functions is to put the function in y = mx + b form, then find 3 ordered pair solutions. In this example, the x-values were chosen for convenience, since the fraction will cancel. You need a minimum of two points to construct the graph of a line, 3 points are used for safety. Another approach to graphing includes computing x and y intercepts, then plotting just those points. You find an x-intercept by setting y=0 and solving for x. The opposite is true for the y-intercept.
In addition to these methods, you should be able to work with linear equations that represent horizontal and vertical lines. The equations of these lines are y = k and x = k respectively, where k is some number. Application Problems You should be able to work with simple application problems that describe a linear relationship. For instance, Fahrenheit and Celsius temperatures are related by the following formula: 5F - 9C = 160. Give a measurement in Fahrenheit (for instance, 68 degrees F), you should be able to compute the corresponding Celsius measurement. Also, you should be able to plot a graph of these values, where Fahrenheit lies along the X axis, and Celsius the Y. Slope The notion of slope underpins a lot of mathematics going forward, up to and through calculus. The idea is simple, slope is a numerical measurement of how "steep" a line is. More specifically, from one point to another on a line, slope measures: change in y values change in x values = Rise Run = y y 2 1 x 2 x 1 where (x 1, y 1 ) and ( x 2, y 2 ) are two points on the line Common Mistake: x 2 does NOT mean x squared, it means "the x-coordinate of point # 2". So, you should understand, intuitively, what slope means: is steeper than so it must have a "greater" slope, as well as being able to compute the value of slope (m) using the slope formula m= y 2 y 1 x 2 x 1 Careful with negative numbers, please!
Understand what a positive and negative slopes indicate (as well as zero and undefined): And be able to graph a linear function based solely on information about the slope and a point on the line. Note: y = m x + b formula indicates that m = slope and b = the y-intercept. Computing Equations of Lines Once you understand slope, you can start figuring out the equation of a line given two pieces of information. Besides the y = m x + b formula, you can use the more powerful y y 1 =m( x x 1 ) For example, given two points (2,4) and (8, 1), you should be able to determine the equation of the line by 1) computing the slope of the line: (you should get -1/2) 2) plugging this answer, along with EITHER (2,4) or (8,1) into y y 1 =m( x x 1 ) Let me plug in the (2, 4): y 4 = 1 2 ( x 2) 3) simplify result to y = mx + b. You should get y = 1 2 x+5 NOTE: if you had plugged in (8, 1) in part 2 above, you would have gotten the same result!
Parallel and Perpendicular Lines Just remember the following when dealing with parallel and perpendicular lines: Parallel lines have EQUAL SLOPES. Perpendicular lines have slopes that are NEGATIVE RECIPROCALS Example: for the line 3x + 2y = 6, it's slope is -3/2 (because you put it into y = mx + b form, right?) Any line parallel has a slope equal to -3/2, and any perpendicular line has a slope of 2/3. That's it! With this information, you should be able to answer questions like: "What is the equation of the line perpendicular to 3x + 2y = 6 that passes through the point (-12, 6)?" Answer: y= 2 3 x+14 Section 3.7 This section deals with linear inequalities in two variables. You definitely need to view the videos for this topic, since it will probably be quite new to you. This topic gets revisited in section 4.5. The concept is that there are a LOT of solutions to something like x + 3y > 6. In fact, if you graph x + 3y = 6 (and use a dashed line because it's >, and not ), you'll see that one "side" of this line solves this inequality, and the other side doesn't. You can make this determination by plugging the point (0,0) into the inequality: x + 3y > 6 0 + 3(0) > 6 FALSE! and noticing the result is FALSE. Since it's FALSE, (0, 0) doesn't "work", and neither do any of its immediate neighbors. So it's the other side of the dashed line that solves this inequality! Sections 4.1, 4.2 Here, we solve systems of inequalities by various methods: graphing, substitution, and addition. Typically, two lines intersect at a single point. This is called an independent system. However, it's easy to see that there may be systems where the lines never intersect (parallel lines - inconsistent system), or overlap each other (infinitely number of solutions - all the points on either line - a dependent system). IMPORTANT: Algebraically, most of the time, you'll get an independent system ( a single solution). However, if you solve your system and the variables CANCEL EACH OTHER OUT, then the resulting values will clue you in to whether the system is inconsistent or dependent. Here's an example:
Starting with: Starting with: 3x + 2y = 8 3x + 2y = 8-3x - 2y = 12-3x - 2y = -8 you'll do some work and determine that all your variables disappear, leading to something like this: you'll do some work and determine that all your variables disappear, leading to something like this: 0 = 20, which is NEVER true 0 = 0, which is ALWAYS true (how can zero equal 20???) (when does zero NOT equal itself???) so the solution is NO SOLUTION or inconsistent system! so the solution set is INFINITE, or a dependent system. Note: I'm not going to ask you to express dependent systems in the form (x, -3x + 4), see Example 6, section 4.1. If the system is dependent, that's all you need to tell me. For algebraic solution methods, we'll cover two of them: substitution, where you plug one of the equations into the other, and addition, where you add together MULTIPLES of the equations together to ELIMINATE one of the variables! That's the key, to reduce the system of 2 equations and 2 variables down to a single equation with one variable! Example of Addition method: 2x + 3y = 11 4x - 2y = -2 If you just "add" these as-is, you would get 6x + 1y = 9. This is NO GOOD, because you reduce the number of variables. However, if you multiply the first equation by -2: 2x + 3y = 11 becomes -4x - 6y = -22, and then add it to the second equation: -4x - 6y = -22 4x - 2y = -2 ----------------- 0-8y = -24 (notice the x's are GONE)... solve for y: y = 3, figure out x from either original equation. So the solution set here is x = 1, y = 3, or (1, 3). 3 by 3 Linear Systems These systems (at the end of section 4.2) look hard, but in fact you're going to follow the same principal you did before... creatively add together multiples of the original equations in such a way that the same variable gets eliminated twice. In other words: reduces a 3 by 3 system down to a 2 by 2 system. Since you know how to solve 2 by 2 systems already, you should be all set.
Example: x + 2y - 2z = 8 (equation 1) 2x + 3y + 2z = 2 (equation 2) x - y - 2z = -6 (equation 3) The easiest variable to cancel twice is "z", since the coefficients are all similar. In fact, if you add equation1 + equation2 (3x + 5y = 10) then add equation2 + equation3 (3x + 2y = -4), you now have a nice, easy 2 by 2 system to solve. 3x + 5y = 10 3x + 2y = -4 I'll let you work through the rest of the problem. Again, view the videos! Systems of Linear Inequalities If you master the material in section 3.7, then this section is trivial. The idea is that we have a system of 2 linear inequalities. Simply solve each inequality separately, then determine the overlapping region. Quick example. Note the overlapping region on the right hand graph. That's the solution set! Exponential Expressions A lot of material here, but much, if not all, should be review. In the first section, it's crucial that you know your rules of exponents! Practice lots of problems. An expression is not considered simplified until: all bases are combined as much as possible, so x 3 y 2 x simplifies to x 4 y 2 no negative exponents, so x 5 y 3 5 z 2 should become x 5 z 2 5 y 3 no parentheses, so ( 4 x 2 y 3 ) 2 should become y 6 16 x 4
One of the applications of working with exponents and monomials in section 5.1 is scientific notation. You've probably done some work with scientific notation in the past, so again, this is probably a review. You should be able to convert to, and from, scientific notation. Polynomials You need to be able to add, subtract, and multiply polynomials for this test. The idea of adding and subtracting is to 1) remove parentheses, and 2) combine like terms. For adding and subtracting, I don't care if you use a horizontal or vertical format, either way is fine. When subtracting polynomials, make sure you add the OPPOSITE of the second polynomial: (3 x 2 5 x+2) (6 x 2 7 x 9) = 3 x 2 5 x+2 + 6 x 2 +7 x+9 Multiplication of polynomials is handled by the FOIL method for binomial time binomial, and extended appropriately for larger polynomials (binomial times trinomial). There are lots of examples in the videos, practice up! Finally, don't get fooled by something like this: (x +2) 2... the answer is NOT x 2 +2 2. The square of a polynomial is that polynomial, times itself! So: (x+2) 2 = (x+2)( x+2) = x 2 +2 x+2 x+4 = x 2 +4 x+4 Good luck with this module!