Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities

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1 Section P.9 Notes Page P.9 Linear Inequalities and Absolute Value Inequalities Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers might be the answer. In this section we will discuss the different ways to write the answers to such problems. The table below is from the tet and eplains the different types of intervals. Besides writing out the intervals, these problems will require you to represent your answer on a number line. Parenthesis indicate that the endpoints are NOT included in an interval. Square brackets indicate the that endpoints ARE included on the interval. Whenever the interval ends with or, parenthesis are always used. That is because or are not eact numbers.

2 Section P.9 Notes Page The table below from the tet lists the nine possible types of intervals used to represent different types of answers. You will notice that there are two types of ways to represent the answer. The first one is called setbuilder notation. This answer is in the form of a set, hence the { and } notation. Your answer always begins with a { and then you write the statement and close it with a }. Then there is interval notation. This is where you use the brackets and parenthesis as discussed on the previous page. The smallest number always comes first in interval notation. In the following problems, epress each interval in set-builder notation and graph the interval on a number line. EXAMPLE: [ 5, ) For this problem, we will use the table above. We look at the interval notation that was given, and match it to the form on the table. We see that -5 would be the a. This states our set-builder notation would be: { 5}. Using the table, we can also epress this answer on a number line. The table above states that our graph will start with -5 and have an arrow pointing to the right. Therefore our correct number line graph is: 5 EXAMPLE: (, ) Using the table again, it says that the following is the correct set-builder notation: { }. Using the table, we can also epress this answer on a number line. The table above states that our graph will start with and have an arrow pointing to the left. Therefore our correct number line graph is:

3 EXAMPLE: [4, ) Section P.9 Notes Page Using the table again, it says that the following is the correct set-builder notation: { 4 }. Using the table, we can also epress this answer on a number line. The table above states that our graph will be between 4 and. There is a bracket on 4 and a parenthesis on the : 4 In a previous section we covered solving linear equations. Now we will solve linear inequalities. To solve these, just pretend like the inequality symbol is an equals sign, and isolate the variable like was done previously. The difference is now we will represent our answer using interval notation and a number line. EXAMPLE: After we isolate, now we need to write this in interval notation using the table this would be written as: [, 9] _ The number line would look like this: EXAMPLE: 4( ) Notice that to solve for, I needed to divide by 7 in order to get a positive What you also noticed was the inequality sign changed directions. This is a rule Whenever you multiply or divide an inequality by a negative, the inequality _ sign will ALWAYS flip. It will not flip with addition or subtraction Interval notation: ( 4, ). Number line: 4 4 EXAMPLE: For fraction problems like this, multiply both sides by the LCD We write the LCD net to each fraction ( 8) (0 ) 5( ) Here we reduced and the fractions were eliminated Add like terms on the left side of the equation _ 44 Notice that this is not a true statement, so the answer is NO SOLUTION.

4 Solving Compound Inequalities Section P.9 Notes Page 4 These are different than the linear inequalities because now you have three sides of an equation. For these, you will do the same operations as in linear inequalities, however you will do operations to all three sides of the equation. Solve the following inequalities and epress the answer in interval notation and on a number line. EXAMPLE: Here I am subtracting 5 from all three sides of the equation to isolate. I am dividing all three sides by to get a single. 4 Now the has been solve for. I now need to write this in interval notation and a number line. I will use the same table I used previously to write the answers. The interval notation is: (, 4). I want you to notice that this is NOT a coordinate like (, y). This is an interval that does not include the or the 4. Now I will epress the answer using a number line: 4 EXAMPLE: I will multiply everything by to cancel out the fraction. 8 6 Everything was multiplied by, so now the fraction is gone since / = I need to divide everything by so that the is positive. Doing this will flip the inequality sign. Remember whenever you multiply or divide by a negative it flips. 0 4 All I did here was flip the inequality so that the smaller number comes first. Notice that the inequality still opens up towards the 0 and the. 4 0 Now we are ready to write the interval notation and the number line using the same table I used previously. The interval notation is: [4, 0). Now I will do the number line: 4 0

5 Solving Absolute Value Inequalities Section P.9 Notes Page 5 Assume that u is any algebraic epression and c is a positive number. Depending on what kind of absolute value inequality you have, you will set up the problem based on:.) If you have u c then you want to solve the following inequality: c u c..) If you have u c then you want to solve the following inequality: c u c..) If you have u c then you want to solve the following inequality: u c or u c. 4.) If you have u c then you want to solve the following inequality: u c or u c. EXAMPLE: In this case we have u c. Therefore we will solve the inequality u c or u c. or Here we set up the inequality. Now we need to solve each one for > or < -9 Now we need to write this with interval notation. Each inequality will be turned into its own interval: (, 9) (, ) The U in the middle is to indicate an or. Now we need to epress our answer as a number line. We will epress each statement on the same number line. It will look like: -9 Remember that numbers that are less than go to the left. Numbers greater than go to the right. EXAMPLE: 4 7. If the absolute value is not isolated, you FIRST must isolate it. We need to get it into the proper form so that we know what equation to solve. Add to both sides. You will get 4 8 In this case we have u c. Therefore we will solve the inequality c u c Here we set up the inequality. Now we need to solve this for Add 4 to all sides of the equation. 4 Now divide all sides by. 4 4 Here is our final inequality. Now we need to write this in interval notation. Interval notatation: 4, 4. Number line answer: 4 4

6 EXAMPLE: 4 8. Section P.9 Notes Page 6 If the absolute value is not isolated, you FIRST must isolate it. We need to get it into the proper form so that we know what equation to solve. Subtract 4 from both sides. You will get 4. Now divide both sides by. You will get: In this case we have u c. Therefore we will solve the inequality c u c. Here we set up the inequality. Now we need to solve this for. Subtract from all sides of the equation. 5 Now divide all sides by since we want to get by itself. 5 Notice here that when we divided by a negative number, the sign switched directions. This will always happen whenever you multiply or divide an inequality by a negative. 5 All I did here was flip the inequality so that the smaller number comes first. Notice that the inequality still opens up towards the 5 and the. Interval notatation: [, 5]. Number line answer: 5 EXAMPLE: Solve: 4 0 Be careful with this one. Remember that an absolute value will ALWAYS return a positive value. Since you will always get a number that is either zero or greater than zero, then this problem has infinite solutions,. Any number will work. The graph would be a number line with everything shaded. EXAMPLE: Solve: 4 0. For this one the only solution is when it equals zero, so you would only solve the equation 4 0, so is 4/. EXAMPLE: Solve: 4 0 Since zero is not included this would have no solution. Since an absolute value always returns a number 0. EXAMPLE: Solve: 4 0 In this case we have u c. Therefore we will solve the inequality 4 0 or 4 0. Solving this 4 4 would give you the following: or. This is saying the same thing as number is included as the answer EXCEPT four-thirds. 4. Therefore every

7 . Graphs and Graphing Utilities Section. Notes Page Cartesian Coordinate System This is the standard graphing system we will use in this course for graphing all kinds of equations. The graph is made up of four quadrants as shown below: The vertical ais is called the y ais and the horizontal ais is called the -ais. The place where the two aes come together is called the origin and the coordinates are (0, 0). We write points in the form (, y). A point is also referred to as an ordered pair. The first number represents the horizontal change, and the second number y represents the vertical change. You move to the right for positive values and to the left for negative values. You move up for positive y values and down for negative y values. Plotting points To plot a point, start at (0, 0). Then move left or right depending on the value and up or down depending on the y value. EXAMPLE: Plot (-, 4). First we start at (0, 0). Since the value is negative we will first move place to the left. Then from this spot we will move up 4 places since the y-value is positive. EXAMPLE: Plot (-4, -). First we start at (0, 0). Since the value is negative we will first move 4 places to the left. Then from this spot we will move down places since the y-value is negative. EXAMPLE: Plot (, -). First we start at (0, 0). Since the value is positive we will first move places to the right. Then from this spot we will move down places since the y-value is negative. EXAMPLE: Plot (0, -). First we start at (0, 0). Since the value is zero we will not move in either direction. We will stay on the y-ais. Then from this spot we will move down places since the y-value is negative.

8 Graphing Equations by Making Tables and Plotting Points Section. Notes Page The equations we are graphing in this section will be done by making a table of values and then plotting the resulting points. Usually the question will tell you which values to use for ; otherwise use whichever values for you want. You should plot negatives, positives, and the zero. Later in this course we will learn how to graph these without making tables. EXAMPLE: Graph the equation y. Let = -, -, -, 0,,,. Then indicate the intercepts. It doesn t matter if you ve never graphed an absolute value before. We plot this one the same way we plot lines. First we need to set up a table like this: y (, y) Now what we do is plug in the value into the equation y. Then we write our answer as an ordered pair as show below. Remember that absolute values always return a positive value. For eample 5 5. y (, y) - y ( ) 4 4 (-, -) - y ( ) (-, -) - y ( ) 0 (-, 0) 0 y (0) (0, ) y () 0 0 (, ) y () (, ) y () 0 (, 0) To get the graph, just plot all these points. You will get this: Let s talk about the term intercepts. This is where the graph crosses either the vertical or horizontal ais. Where ever the graph crosses the vertical ais is called the y-intercept. Where ever the graph crosses the horizontal ais is called the -intercept. The problem asked us to identify the intercepts. This means we need to read them off our graph. For the y-intercept, this point would be (0, ). There are two places where it crosses the horizontal ais, and this occurs at (-, 0) and (, 0).

9 Section. Notes Page EXAMPLE: Graph the equation y 4. Let = -, -, -, 0,,,. Then indicate the intercepts. It doesn t matter if you ve never graphed a quadratic equation before. We plot this one the same way we plot lines. First we need to set up a table like this: y (, y) Now what we do is plug in the value into the equation y 4. Then we write our answer as an ordered pair as show below. Remember that a negative number r aised to an even power is positive. For eample ( 5) ( 5)( 5) 5. y 4 (, y) - y ( ) (-, 5) - y ( ) (-, 0) - y ( ) 4 4 (-, -) 0 y (0) (0, -4) y () 4 4 (, -) y () (, 0) y () (, 5) To get the graph, just plot all these points. You will get this: Now let s identify the intercepts the same way we did it for the previous problem. The y-intercept is where the graph crosses the vertical ais. This point would be at (0, -4). There are two places where it crosses the horizontal ais, and this occurs at (-, 0) and (, 0). You can also write this as:, 0. Graphing Utilities This book does some eercises where it wants you to set up viewing windows. Since I do not require graphing calculators in this class, I am not requiring you to do these types of eercises. If you already have a graphing calculator, you may use it for this class. Please ask me after class or in office hours if you have a specific question on how to use your particular graphing calculator.

10 . Basics of Functions and Their Graphs Section. Notes Page Domain: (input) all the -values that make the equation defined Defined: There is no division by zero or square roots of negative numbers Range: (output) all y-values that a graph uses. EXAMPLE: Find the domain and range of the following graph (assume graph ends at edge of graph and the bottom edge of the graph is the -ais, and the left edge of the graph is the y-ais) Domain: [0, 4] Range: [0, ] Function Definition: For each input () there can only be one output (y). EXAMPLE: For each relation below, determine whether it is a function. Then give the domain and range for each relation. {(, ), (, 7), (, 9), (8, )} This is a function. Every goes to only one y. The domain of this is all the -values. The answer is {,,, 8}. You may leave it this way or order them. The range of this is all the y-values. The answer is {, 7, 9, }. {(-, 4), (5, 6), (7, 4), (-, )} This is a function. Even though the -values - and 7 both go to 4, each value goes to only one y-value. Domain: {-, 5, 7, -} Range: {4, 6, } Notice that 4 is repeated, but you only need to write it once. {(-, 4), (-, 6), (0, ), (-, 8)} NOT a function because when is - it goes to both 4 and 8. There are two different y values for one. Domain: {-, -, 0} Notice again that even though - is repeated, it only needs to be written once. Range: {4, 6,, 8} {(5, ), (-, ), (5, ), (9, 0)} This is a function. The same point repeats, but still goes to only one y. Domain: {5, -, 9} Range: {,, 0}

11 Section. Notes Page Vertical line test. If you pass an imaginary vertical line through the graph and it only intersects the graph once then it is a function. Which graphs below are functions? Function NOT a function. Function EXAMPLE: Is y 7 a function? We don t have a graph drawn for us or a set of points. We need to see if it is a function algebraically. First think we need to do is solve for y. We will isolate it and then take the square root of both sides. Don t forget that you will get a plus and minus whenever you take the even root of something. y 7 y 7 7 y Notice that for each we will get two different y values because of the. Therefore we know this is not a function. EXAMPLE: Is y 5 a function? Again we will solve for y. When we do we will take the odd root of both sides. There will be no plus and minus here since it was an odd root. y y 5 5 To get rid of the fifth power, I took the fifth root. For each we put in we will only get one y-value for each we put in so it IS a function. So what is the general rule here based on our previous two eamples? Any equation that has a y raised to an even power is NOT a function. Any equation that has a y raised to an odd power IS a function.

12 Section. Notes Page Function notation: f () which means f of. This does not mean f times. It means that we have a function called f which contains the variable. EXAMPLE: Given the function f ( ) 5, find the following: a.) f () Whatever is inside the parenthesis goes in place of in the original epression. This is really asking us for the y value when is. f ( ) () 5 f ( ) b.) f ( ) Now we need to replace in the original equation with +. Then simplify. f ( ) ( ) 5 f ( ) 6 5 f ( ) This is as far as we can go on this one. c.) f ( ) f () For this one we can replace the f () with 5. We also know f (). f ( ) f () 5 f ( ) f () 4 Notice this is not the same as part b, so the f is not distributed to the and. d.) f ( h) For this one just replace the with the epression + h. f ( h) ( h) 5 f ( h) h 5 This is as far as we can go. EXAMPLE: Let a.) f (5) f ( ) 4. Find the following: 5 4 f (5) We are replacing with 5. (5) f ( 5) 7

13 b.) f ( h) Section. Notes Page 4 ( h) 4 f ( h) We are replacing with the quantity ( + h). ( h) h 4 f ( h) This is as far as we can go. h c.) f ( ) f (5) 4 7 We are replacing f() with our original function and f(5) we found in part a Generally if you have two fractions, then combine after common denominators. 7( 4) ( ) 7( ) Now add the fractions together now that we have common denominators ( ) Distribute and simplify. 9 7( ) This is our final answer. EXAMPLE: Given f ( ), find f ( ). f ( f ) ( ) ( ) Replace with and simplify. ( ) We have looked at function notation for equations, but now we will see the relationship between the function notation and graphs. This net eercise shows how to read values off a graph, but first let s talk about symmetry. Symmetry: -ais symmetry y-ais symmetry Origin Symmetry (-ais is a fold line) (y-ais is a fold line) (Graph is in opposite quadrants)

14 EXAMPLE: Use the graph below to answer the following: Section. Notes Page 5 a.) Find the domain Since we don t include the endpoints we have (-, ) ( values) b.) Find the range The answer is (-, ] (y-values) c.) Indicate the intercepts -int: (-, 0) (, 0) y-int: (0, ) EXAMPLE: Use the graph below to answer the following: d.) Indicate any symmetry this graph has. You can fold this in half over the y-ais, so it has y-ais symmetry. a.) Find f (): This is asking you for the y value when is -. The answer is f () =. b.) Find all such that f ( ) This is asking you to find all that give a y value of. This happens at the point (5, ), so = 5. c.) Is f ( ) positive or negative? This is asking you if the y value at = is above or below the ais. To find this go over to =. We notice the graph is below the -ais, so answer is neg. d.) What is the domain? This is asking you for all the values the graph uses. This would be [-4, 6]. (lowest to highest ) e.) What is the range? This is asking you for all the y values the graph uses. The answer is [-, ]. (lowest y to highest y). f.) For which values is f ( ) 0? This is asking you which part of the graph has positive y values. In other words, what part of the graph is above the -ais, but not on the -ais. We have two places this occurs. [-4, 0) or (4, 6) Notice the values I gave in the interval notation are values. We include the -4 because it is not on the -ais.

15 EXAMPLE: Use the graph below to answer the following: Section. Notes Page 6 a.) Find f (): This is asking you for the y value when is -. The answer is f () =. It does not matter if the -value has a dot or not. b.) Find all such that f ( ) 0 This is asking you to find all that give a y value of 0. This happens at =,, and 5. c.) Is f positive or negative? This fraction is the same as -.5. When you go to this value the graph is above the -ais here, so positive. d.) What is the domain? The domain is referring to the -values the graph uses. Since there is an open circle at, this -value is not included. So the domain is: (, 6]. e.) What is the range? The range is referring to the y-values the graph uses. Again since there is an open circle at, this y-value is not included. So the range is ( 4, 4]. f.) Indicate the and y intercepts. y-int: (0, ) -int: (-, 0), (, 0), (5, 0). g.) Indicate what kind of symmetry, if any, this graph has. This graph does not have any symmetry.

16 . More on Functions and Their Graphs Section. Notes Page Even and Odd functions If If f ( ) f ( ) then the function is even, and symmetric to the y-ais. f ( ) f ( ) then the function is odd, and symmetric to the origin. EXAMPLE: Determine whether the following are even, odd, or neither. a.) f ( ) 4 7 We want to put a in for first. You will get: f ( ) ( ) 4 7 This simplifies to f ( ) 4 7. Since this is the same as the original, this function is even. b.) f 5 ( ) 6 5 We want to put a in for first. You will get: f ( ) 6( ) ( ) 5 f ( ) 6 function is odd. This simplifies to 5. Factoring out a negative: f ( ) (6 ). So we have f ( ) f ( ) so this c.) f ( ) We want to put a in for first. You will get: f ( ) ( ) ( ) This simplifies to f ( ). There is nothing more I can do to this one. First we notice this is not the same as the original so it is definitely not even. If we try to factor out a negative we get f ( ) ( ). Since you don t have f() inside of the parenthesis it is not odd either. We will answer neither. d.) f ( ) If we put in a for we will get: function will be odd. ( ) f ( ) which simplifies to f ( ) which means the Increasing: as increases, y increases (graph goes uphill as you move from left to right) Decreasing: as increases, y decreases (graph goes downhill as you move from left to right) Constant: as increases, y does not change (this part of the graph is horizontal) Relative maimum: a point at which the graph increases and then decreases (peak) Relative minimum: a point at which the graph decreases and then increases (valley)

17 Section. Notes Page EXAMPLE: Use the graph below to answer the following questions a.) Indicate the interval(s) of which f is increasing There are two places this occurs. Between the values of - and 0 the graph is climbing. This also happens between,0, 4. W use parenthesis because at the endpoints the graph is not increasing or decreasing. and 4. We write the answer as e always b.) Indicate the interval(s) of which f is decreasing There is one place where the graph is falling as we move from left to right. This is between the value of 0 and, so we write our answer as (0, ). c.) List the number where f has a relative maimum. This is asking for the value at which the graph has a local maimum. This occurs at = 0. d.) What is the relative maimum? This is asking for the y-value of the local ma, which is. e.) What is the relative minimum? The y-value of the local minimum is 0. EXAMPLE: Use the graph below to answer the following questions a.) Indicate the interval(s) of which f is increasing (,0) (,5) b.) Indicate the interval(s) of which f is decreasing (, ) (0,) c.) List the number(s) where f has a relative minimum. This is asking for the value(s) at which the graph has a local minimum. This occurs at = - and at =. d.) What is the relative maimum(s)? This is asking for the y-value of the local ma, which is -. e.) What is the relative minimum(s)? The y-value of the local minimum is - and -5. f.) What is the domain? [, 5) g.) What is the range? [-5, 0]

18 Piecewise Functions Section. Notes Page These functions are made up of different pieces. Each piece is defined for certain values of. if EXAMPLE: Use the function f ( ) to find f (4), f () and f. Then graph. if and use this to determine the graph s range. a.) f (4) In order to know which equation we are using, look at the number inside the parenthesis, which is 4. In our function, we need to find what function includes -4. This would be the first equation since -4 is less than -. So we put -4 in for in the first equation. You will get -4 + = -. So f ( 4). b.) f () The equation that includes - would be the second one since it is greater than or equal to. So we will place the - in for in the second equation: - = -5. So f ( ) 5. c.) f If you don t know if this fraction is less or more than - then turn it into a decimal. This is -.5. So this would be greater than -, so we will use the second equation So f. Now we do a graph. Notice that there are two different equations we need to graph. To graph these, we will make a table of values for each one. The most important thing is that we need to plug in values that match our conditions in the problem. For eample the first equation says we need to use values that are less than but not equal to -. Now we can plug in -, and this will end up as an open circle on the graph since it is not included. So we can use = -5, -4, -. Three points are enough. For the second equation we need to pick values that are greater than or equal to -. When we plug in - we can plot this point as a closed circle since this point is included. We will use = -, -, -. Below are a table of values for each equation. To graph this, we plot points from our table making sure to indicate a closed or open circle: y (, y) -5 y 5 (-5, -) -4 y 4 (-4, -) - y (-, -) y (, y) - y 5 (-, -5) - y 4 (-, -4) - y (-, -) Notice the open circle at (-, -). Also notice that there are no values plotted to the right of the open circle because this graph is only defined for values less than or equal to -. Notice the open circle at (-, -5). Notice that there are not values plotted to the left of this point since we are only allowed to use values for that are greater than or equal to -. The range is the y-values the graph is using. This would be ALL y values, so the answer is,.

19 if EXAMPLE: Use the function f ( ) to find f (), f () and if use this to determine the graph s range. f Section. Notes Page Then graph. and a.) f () In order to know which equation we are using, look at the number inside the parenthesis, which is -. In our function, we need to find what function includes -. This would be the first equation since - is less than. So we put - in for in the first equation. You will get ( ) (). So f ( ). b.) f () The equation that includes would be the second one since it is greater than or equal to. So we will place the in for in the second equation: () + = + =. So f ( ). c.) 9 f If you don t know if this is less or more than then turn it into a decimal. This would be 0.95, 0 which is than, so we will use the first equation So, 9 9 f 0. 0 Now we do a graph. Notice that there are two different equations we need to graph. To graph these, we will make a table of values for each one. The most important thing is that we need to plug in values that match our conditions in the problem. For eample the first equation says we need to use values that are less than but not equal to. Now we can plug in, and this will end up as an open circle on the graph since it is not included. So we can use = -, 0,. Three points are enough. For the second equation we need to pick values that are greater than or equal to. When we plug in we can plot this point as a closed circle since this point is included. We will use =,,. Below are a table of values for each equation. To graph this, we plot points from our table making sure to indicate a closed or open circle: - 0 (, y) y y ( ) (), (0, 0) y (0) 0 y (), y (, y) y () (, ) y () 5 (, 5) y () 7 (, 7) The range is the y-values the graph is using. The graph is not using any y value between 0 and, so we write, 0,. our answer this way:

20 Section. Notes Page 5 0 if 0 EXAMPLE: Use the function f ( ) if 0 to find f (0), f (), f, f ( ). if 0 a.) f (0) Zero is only included in the third equation. We put a zero in for in the third equation: 0. So we write f ( 0) b.) f () Negative one is included in the middle equation since it is between - and 0. We will put in - for. We will get: ( ). So we write f ( ). c.) 0 f Turning it into a decimal we get.58, so we use the third equation this time. So f. 4 d.) f ( ) Turning this into a decimal we get -., so we use the first equation. The answer is 0. Now we do a graph by making a table of values for each equation. Notice that there are three different equations we need to graph. To graph these, we will make a table of values for each one. For the first equation we use the following values:,,. For the second equation we can use = -, -, 0. For the third equation we can use = 0,,. y 0 (, y) y 0 (, 0) y 0 (, 0) y 0 (, 0) y (, y) - y ( ) (-, ) - y ( ) (-, ) 0 y ( 0) 0 (0, 0) y (, y) 0 y (0) (0, -) y () 0 (, 0) y () (, ) Notice that our first equation is a horizontal line with an open circle at (-, 0). Notice an open circle at (0, 0), and closed circle at (0, -). The range is the y-values the graph is using. The graph is not using any y values less than, so we write our answer as:,.

21 Difference Quotient Section. Notes Page 6 If we wanted to find the slope of a curved line, the only way we can do this is by estimating it with a straight line. We will start with one point and then move over by a small amount h. Now we will use the slope formula. In the picture we have two points, A and B. The coordinates for these are: (, f ( )) and ( h, f ( h)). f ( h) f ( ) The slope, also called the difference quotient is: h In calculus we will try to minimize h so that it is so small that we end up at a point, which will be the eact slope of the curved line at. Now let s look at some eamples finding the difference quotient. EXAMPLE: Let f ( ). Find the difference quotient. Let s first find f ( h). Once we have this we can put it into the difference quotient formula. Replace in the original equation with + h. f ( h) ( h) Now simplify. f ( h) h We are ready to substitute this into difference quotient formula. We have f ( h) and we also know f (), which is the original equation. h ( ) Here we have substituted into the formula. Notice the parenthesis around f (). h h Now we distributed the minus sign and the last thing is to simplify. h h The and the canceled and then the h canceled, leaving us with our answer of. h Does make sense? Yes, because a difference quotient tells you the slope at any value of. Since we have a line the slope of will not change no matter what value of we use. EXAMPLE: Let f ( ). Find the difference quotient. We will do this the same way as above. First we will find f ( h). f ( h) ( h) ( h) What is ( h)? If you are thinking h you are wrong. This is actually ( h)( h) which is a FOIL. It is h h. f ( h) ( h)( h) h f ( h) ( h h ) h f ( h) 6h h h Now that we have simplified this as much as possible, we will put it into the difference quotient formula.

22 Section. Notes Page 7 6h h h ( ) h Now we will distribute the minus into f (). 6h h h h Now cancel and simplify. 6h h h h Now we can factor out an h from the top. h ( 6 h ) h Last thing is we can cancel the h from top and bottom. 6 h This is our answer. What can you do with this epression? Now if we were in calculus we would minimize the h (go to zero). What is left can be used to find the slope of this curve at any point (derivative). EXAMPLE: f ( ). Find the difference quotient. f ( h) Nothing more we can do to simplify this. Now put it into the formula. ( h) h Now to simplify this we need add the fractions on the top. Need common denominators. h ( ) ( h ) ( ) h ( h ) Now that we have common denominators, combine the fractions h ( ) ( h ) ( )( h ) We can simplify the very top part of this fraction h h ( )( h ) Simplify. h h ( )( h ) Now we will clear the double fractions by multiplying by the reciprocal. h h h( )( h ) ( )( h ) Almost done. Just need to cancel out the h from the top and bottom. Whew! Okay this is our answer.

23 .4 Linear Functions and Slope Section.4 Notes Page This section is designed to be a review from Intermediate Algebra. Slope Formula The slope formula is used to find the slope between two points, y and, y. (, y ) The slope is the vertical change divided by the horizontal change. From our picture, the horizontal change is y y and the horizontal change is. (, y) From this we get the formula for slope: m y y. Positive slopes will rise as you move from left to right. Negative slopes will fall as you move from left to right. A slope of zero is a horizontal line. An undefined or infinity slope is a vertical line. EXAMPLE: Find the slope of a line passing through the following points. Indicate whether the line rises, falls, is horizontal or vertical. a.) (-, ) and (, 4) To do this problem we can label our point so we know what to put into the slope formula. It doesn t matter which point you call or. I will label the point as the following:, y,, y 4. Now 4 we plug these into the slope formula: m. Since the slope is positive we know this line rises. ( ) b.) (4, -) and (, -) First we label the point as the following: 4, y,, y. Now we plug these into the slope ( ) 0 formula: m 0. Since the slope is zero we know this line is horizontal. 4 c.) (, -) and (, -5) First we label the point as the following:, y,, y 5. Now we plug these into the slope 5 ( ) formula: m undefined. Since the slope is undefined we know this line is vertical. 0

24 Now I will introduce two formulas you will need to know in this section: Section.4 Notes Page Slope-Intercept Formula this is the standard from of a line which allows you to easily identify the slope and y-intercept. y m b Here the slope is m and the y-intercept is (0, b). Point-Slope Formula this is used when you want to find the equation of a line when you are give a slope and another point on the line. This other point does not need to be the y-intercept. y y m( ) EXAMPLE: Use the information and given conditions to write an equation for each line in slope-intercept form as well as the point-slope form. a.) Slope = 8, passing through (4, -). For this one, we know that m = 4, 4, and y. We can plug these into our point-slope formula: y ( ) 8( 4). When we simplify we get: y 8( 4). The equation of this line is now written in slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form. To do this, we just need to solve for y. First we distribute the 8: y 8. Now subtract from both sides to get our second answer: y 8. b.) Slope =, passing through (0, -4). 5 For this one, we know that m =,, and 5 0 y 4. We can plug these into our point-slope formula: y ( 4) ( 0). When we simplify we get: y 4 ( 0). The equation of this line is now 5 5 written in slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form. First we distribute the : 5 0 y 4. To multiply the two fractions on the end, multiply 5 5 across the top and bottom. You will get: y 4 6. Now subtract 4 from both sides to get our second 5 answer: y. 5

25 c.) Passing through (-, 6) and (, -) Section.4 Notes Page This time we are not given a slope, so we first must use the slope formula. We label our points and put them into the slope formula: m. When we use the point-slope formula we can use EITHER ( ) 6 of our given point as the (, y). In this case I will use the first point. So, and y 6. We can plug these into our point-slope formula: 4 4 y 6 ( ( )). When we simplify we get: y 6 ( ). The equation of this line is now written in slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form. First we distribute the : y 6 4. Now add 6 to both sides to get our second answer: y. d.) -intercept =, y-intercept = 4 This time we are not given a slope, so again we must use the slope formula. We want to put these intercepts in a point form, which would be:, 0 and 0, 4. We label our points and put them into the slope formula: m 8. When we use the point-slope formula we can use EITHER of our given point as the 0 (, y). In this case I will use the second point. So 0, and y 4. We can plug these into our pointslope formula: y 4 8( 0). When we simplify we get: y 4 8. The equation of this line is now written in slopeintercept form, which is one of our answers. Now we need to write it in slope-intercept form. We just need to add 4 to both sides to get our second answer: y 8 4. EXAMPLE: Write the following in slope-intercept form and identify the slope and y-intercept. Use this information to graph the equation. a.) f ( ) 4 This equation can be written in slope-intercept form by replacing From here we can identify that the slope is 4 and the y-intercept is (0, -). To graph this, first plot the y-intercept. Now we need another point on our line. The definition of the slope is the change in the vertical distance divided by the change in the horizontal distance. In our slope the top number is. This is our vertical change. Because it is positive we will move up units from our y-intercept. The bottom number is 4, so we will need to move 4 units to our right. So from our y-int we will move up units and 4 units to the right. This will give us our net point. Plot this and connect our two points with a line. f () with y. You will get: y. 4

26 b.) 4 + 6y + = 0 Section.4 Notes Page 4 We need to solve for y in order to put this into slope-intercept form. First isolate y: 6y = -4. Now divide both sides by 6 to get: y. Now we can identify that the slope is and the y-intercept is (0, -). To graph this, first plot the y-intercept. Now the fraction can be written as either or. If we think of the slope as then the our vertical change is -. This means we move DOWN units from our y-intercept. The bottom number is, so we will need to move units to our right. So from our y-int we will move DOWN units and units to the right. This will give us our net point. Plot this and connect our two points with a line. If we used then we would move UP two units and to the LEFT units. Notice we will still get another point on the same line, so we can use either fraction. EXAMPLE: Find the and y intercepts and use them to graph the following equation: 6 + 9y 8 = 0. To find an -intercept, put a zero in for y and solve: 6 + 9(0) 8 = 0. Simplifying will give us 6 8 = 0. Solve for : 6 = 8, so =. It is important to write our answer as a point. The -intercept is (, 0). To find an y-intercept, put a zero in for and solve: 6(0) + 9y 8 = 0. Simplifying will give us 9y 8 = 0. Solve for y: 9y = 8, so y =. It is important to write our answer as a point. The y-intercept is (0, ). To graph, just plot each point. EXAMPLE: Find the and y intercepts and use them to graph the following equation: 6 y + 5 = 0. To find an -intercept, put a zero in for y and solve: 6 (0) + 5 = 0. 5 Simplifying will give us = 0. Solve for : 6 = -5, so =. 5 It is important to write our answer as a point. The -intercept is, 0. To find an y-intercept, put a zero in for and solve: 6(0) y + 5 = 0. Simplifying will give us -y + 5 = 0. Solve for y: -y = -5, so y = 5. It is important to write our answer as a point. The y-intercept is (0, 5). To graph, just plot each point. For fractions, you can change them into a decimal. Our -intercept can be written as: (-.5, 0).

27 .5 More on Slope Section.5 Notes Page Parallel lines have the same slope. These lines do not cross. Perpendicular lines have opposite reciprocal slopes (opposite sign and one fraction is flipped over). 4 E: and are opposite reciprocals. Also and are opposite reciprocals. Perpendicular lines will 4 cross at a right angle (90 degrees). EXAMPLE: Find the slope of a line that is perpendicular to y = -4. The slope of this line is -4, and because it says perpendicular, we need to find the opposite reciprocal. The 4 number -4 can be rewritten as the fraction. Because it is a negative, the opposite sign will be positive. If we flip over the fraction we get, which is our answer. 4 EXAMPLE: Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. a.) Passing through (-, -7) and parallel to the line whose equation is y 5 4 Since we want a line parallel, then this will have the same slope as our given equation, so we know m = -5. We are also given a point. From here, we will plug this information into the point-slope formula. You should get: y ( 7) 5( ( )). Simplifying gives you y 7 5( ). This is our first answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y Now subtract 7 from both sides: y 5 7. This is our answer in slope-intercept form. b.) Passing through (-4, ) and perpendicular to the line whose equation is y 7 Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation, so we know m = -. We are also given a point. From here, we will plug this information into the point-slope formula. You should get: y ( ( 4)). Simplifying gives you y ( 4). This is our first answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y. Now add to both sides: y 0. This is our answer in slope-intercept form. c.) Passing through (5, -9) and perpendicular to the line whose equation is 7y 0 Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation. We first need to solve our given equation for y: y. The slope of this line is. A line perpendicular to this one will have a slope of m = 7. From here, we will our slope and given point into the point-slope formula. You should get: y ( 9) 7( 5). Simplifying gives you: y 9 7( 5). This is our first answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y Now subtract 9 from both sides: y This is our answer in slope-intercept form.

28 Section.5 Notes Page General Form: A + By + C = 0. In the general form, everything is set equal to zero. The constants, A, B, and C should be written as integers if possible, and A must be written as a positive number. EXAMPLE: Given y 9 7 5, write this in general form. In order to solve this, we must bring everything over to one side of the equation and set it equal to zero. I will move everything from the right side to the left: y Now simplify and write the and y terms first: 7 y The A constant must be written as a positive number, so we will multiply the whole equation by - to get: 7 y Average Rate of Change (A.R.C.) The A.R.C. is an estimate of the slope between and c. Basically how much does something change between and c. The formula is as follows and is derived from the slope formula, much like the difference quotient. y f ( ) f ( ) EXAMPLE: Find the A.R.C. for the f ( ) from 0 to. We can substitute these numbers into our general A.R.C. formula: f () f (0) Now we will work this out. Find f () and f (0) So the slope is -6 between the values of 0 and. EXAMPLE: Find the A.R.C. for the f ( ) from to. We can substitute these numbers into our general A.R.C. formula: f () f () 6 Now we will work this out. Find f () and f () So the slope is between the values of and. EXAMPLE: Find the A.R.C. for the f ( ) from 9 to 6. We can substitute these numbers into our general A.R.C. formula: f (6) f (9) Now we will work this out. Find f (6) and f (9) So the slope is between the values of 9 and

29 .6 Transformations of Functions Section.6 Notes Page We will first look at the major graphs you should know how to sketch: y y y Square Root Function Absolute Value Function Identity Function Domain: [ 0, ) Domain: (, ) Domain: (, ) Range: [ 0, ) Range: [ 0, ) Range: (, ) Increasing: ( 0, ) Increasing: ( 0, ) Increasing: (, ) Decreasing: None Decreasing: (, 0) Decreasing: None y y y Standard Quadratic Function Standard Cube Function Cube Root Function Domain: (, ) Domain: (, ) Domain: (, ) Range: [ 0, ) Range: (, ) Range: (, ) Increasing: ( 0, ) Increasing: (, ) Increasing: (, ) Decreasing: (, 0) Decreasing: None Decreasing: None

30 Transformations and Graph Sketches Section.6 Notes Page When we used to graph a line the usual thing to do was make a table of values and plot the points. This method works but takes a long time. Transformations allows you move a graph up or down, left or right into a new position. We start with the basic graphs we learned in the last section and will move it based on the following criteria. Suppose y = f() is the original function (one we looked at in the previous section) Y = f() + k moves f() k units up Y= f() k moves f() k units down Y = f( - h) moves f() h units to the right Y= f( + h) moves f() h units to the left Y = -f() flips the graph over a horizontal ais Y = f(-) flips the graph over a vertical ais Y = a(f()) If a > then there is a vertical stretch. If 0 a, then there is a vertical compression. Let s look at some eamples. For all of these we are just making a sketch of the function. EXAMPLE: Sketch y by using transformations. First we need to recognize what kind of graph we are using. This is the absolute value graph which is a V shaped graph centered at the origin. Since there is a + inside the absolute value we are looking at rule 4 which says we will move the graph unit to the left. So the graph will look like: What is the -intercept? It crosses the -ais at (-, 0). What is the y-intercept? It crosses the y-ais at (0, ). EXAMPLE: Sketch y by using transformations. This may look similar to the graph above, but it is different because the is on the outside of the absolute values. This is rule which says the graph will move unit up: There are no -intercepts on this one. The y-intercept is (0, ).

31 EXAMPLE: Sketch y by using transformations. Section.6 Notes Page This one has a negative on the outside of the absolute value. Since the negative is outside we will use rule 5 which says the graph will flip over the horizontal ais. This will flip the graph upside down and the pivot point is the origin (0, 0): -int: (0, 0) y-int; (0, 0) EXAMPLE: Sketch y by using transformations. Now let s put it all together. Usually you will be using more than one transformation. This problem does puts the last three eamples together. We start with our absolute value graph at the origin. We will move it one place to the left, then up one unit and then flip it upside down: -intercept: (-, 0) and (0, 0) y-intercept: (0, 0) EXAMPLE: Sketch y by using transformations You will move the graph in the same directions as the last eample, but now there is a number in front of the absolute value. This doubles all of the regular y-values, causing the graph to be vertically stretched. EXAMPLE: Sketch y by using transformations. Now we will look at a different type of graph. This one is the square root. Our transformations now tell us we will move the square root graph places to the right and units down. We don t need to flip the graph because there is no negative outside of the function, or inside net to an.

32 EXAMPLE: Sketch y by using transformations. Section.6 Notes Page 4 This one requires us to move the square root graph places to the left and down one unit. Once this is done we need to flip the graph over the horizontal ais because the negative is on the outside of the equation. EXAMPLE: Sketch y by using transformations. In order to use the transformation rules the must come first and there must be a one in front of. In our problem above we need to first put the first and then we will factor out a negative: y y Here we put the term first y ( ) Here we factored out a -. Now we are ready to graph. Since we already factored out the negative We need to move the graph places to the right and then up units. There is a negative inside the function, so we need to use rule 6 of the transformations which says we will flip the graph over the vertical ais. EXAMPLE: Sketch y by using transformations. Be careful you don t confuse this with the square root graphs we just did. This one looks different. We will move the regular cube root graph to the right unit and up units. EXAMPLE: Sketch y ( ) by using transformations. For this one since the negative is already factored out we will move the graph units to the left and up units. We will also flip the graph over the vertical ais since the negative is inside the radical.

33 EXAMPLE: Sketch y ( ) 4 by using transformations. Section.6 Notes Page 5 Here we have the cube graph. Don t confuse it with the cube root graph. We need to move the graph units to the right and down four units. There is a negative outside so we will flip the graph over the horizontal ais. EXAMPLE: Sketch y ( ) 4 by using transformations. For this one we need to once again put the term first and then factor out the negative sign. y ( ) 4 y ( ) 4 y ( ( )) 4 So we move the graph places to the right and down 4 units. The negative is inside the function, so we need to flip it over the vertical direction. Does this graph look familiar? It is the same graph as the previous problem. Why? Notice the negative inside the function above. If I raise a negative to an odd power then the negative can come outside the parenthesis. If this happens you will get the same equation as the last eample. EXAMPLE: Sketch y ( ) by using transformations. Now we have the squaring function. This will be a parabola. We will move the parabola places to the right and down unit. EXAMPLE: Sketch y ( ( )) by using transformations. This one is shifted in the same directions as in the last eample. We have a negative inside the function which means we will shift it over the vertical ais. If you flip the graph it will look the same as the original, so the graph is the same as the previous eample. Also if you raise a negative to an even power the negative goes away and you will get the same equation as in the previous eample so that is why the graph looks the same.

34 Section.7 Notes Page.7 Combinations of Functions; Composite Functions The following problems deal with finding the domain without a graph, which can be done algebraically. If a function has no places where you are dividing by zero and no places where you are taking the square root of a negative then there are no domain restrictions, so the domain would be all reals. EXAMPLE: Find the domain: y 5 There is no place where you can divide by zero or take the square root of a negative number, so the domain would be all reals, indicated by,. EXAMPLE: Find the domain: y 5 Here it is possible to have a zero in the denominator. The denominator is not allowed to be zero, so solve: Solving this you will get. This means any number but five halves will work. To write this 5 5 in interval notation it would be:,,. EXAMPLE: Find the domain: y 5 For this one you need to make sure you do not take the square root of a negative number. The only numbers that will work are positive numbers, so solve this equation: 5 0. It is okay for our answer to equal Solving it you will get. In interval notation this would look like,. EXAMPLE: Find the domain: y 5 This has two domains restrictions. First the denominator can t be zero. Also we are not allowed to have negative numbers under the square root. We will set it up almost the same as before, but this time we will not include zero. We want to solve: 5 0. We don t want a zero in the denominator, so we don t include it 5 5 in our answer. Solving this we get and the interval notation would be,. EXAMPLE: Find the domain: y The bottom root has an odd inde (little number net to radical, which is ). This is not a square root. Since we have an odd inde that means that if we take the odd root of a negative number, we will get something defined. There for the only domain restriction is if the bottom equals zero, so we solve 0, so. If we wanted to write this in interval notation it would be,,.

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