Supplemental Worksheet Problems To Accompany: The Algebra 2 Tutor Section 4 Graphing Inequalities

Transcription

1 Supplemental Worksheet Problems To Accompan: The Algebra Tutor Please watch Section 4 of this DVD before working these problems. The DVD is located at: Sample Videos For this DVD Are Located Here: Page 1

2 1) Graph the following inequalit: > Page

3 ) Graph the following inequalit: + 1 Page

4 ) Graph the following inequalit: + < 0 Page 4

5 4) Graph the following inequalit: + Page 5

6 5) Graph the following inequalit: > 6 Page 6

7 Question Answer 1) Graph the following inequalit: > Begin. = To graph an inequalit, first graph the line that results when we replace the inequalit with an equal sign. This will be the boundar in our inequalit graph. This, we first graph the equation of the line at left as our boundar. If the inequalit has a > or < sign, we draw this line with a dotted line on the graph because the line itself is not part of the solution. If the inequalit has a or sign, we draw this line with a solid line because the line itself will be a part of the solution. Plug in = 0 = 0 = Point 1 is Plug in =0 0= = ( 0, ) ( 0) Point is, You can use an method ou like to graph to graph the line. For this problem, we choose the intercept method. We plug in =0 and calculate the resulting. We plug in =0 and calculate the resulting. This gives us points on our, plane to draw our line through. (continued on net page) Page 7

8 Plot the two intercept points on our - plane and draw a line through them. Since our original inequalit had a >, we draw this as a dotted line. The line we drew was =. The original inequalit was >. This means that the actual graph that represents this inequalit consists of all points greater than the line = not including the line itself (because of the > sign). You can interpret greater than the line as all points above the line that ou initiall draw. We represent this b shading the graph above the line. Ans: See graph at left. Page 8

9 Question Answer ) Graph the following inequalit: + 1 Begin (subtract from both sides) 1 + (divide both sides b ) + 6 (Simplif) First, alwas write our inequalit with on the left hand side and everthing else on the right hand side. We do this simplification at left. Our goal from here on out is to graph the purple inequalit at left, which is equivalent to what we started with. = + 6 To graph an inequalit, first graph the line that results when we replace the inequalit with an equal sign. This will be the boundar in our inequalit graph. This, we first graph the equation of the line at left as our boundar. If the inequalit has a > or < sign, we draw this line with a dotted line on the graph because the line itself is not part of the solution. If the inequalit has a or sign, we draw this line with a solid line because the line itself will be a part of the solution. (continued on net page) Page 9

10 Plug in =0 = ( 0 ) + 6 = 6 ( ) Point 1 is 0,6 You can use an method ou like to graph to graph the line. For this problem, we choose the intercept method. We plug in =0 and calculate the resulting. Plug in =0 0= + 6 = 6 (add to both sides) We plug in =0 and calculate the resulting. This gives us points on our, plane to draw our line through. (continued on net page) = 1 (multipl both sides b ) = 4 (divide both sides b ) ( ) Point is 4,0 Page 10

11 Plot the two intercept points on our - plane and draw a line through them. Since our original inequalit had a, we draw this as a solid line. The line we drew was = + 6. The original inequalit was + 6. This means that the actual graph that represents this inequalit consists of all points greater than the line = + 6 including the line itself (because of the sign). You can interpret greater than the line as all points above the line that ou initiall draw. We represent this b shading the graph above the line. Ans: See graph at left. Page 11

12 Question Answer ) Graph the following inequalit: + < 0 Begin. + < 0 < 0 (subtract from both sides) < (Simplif) First, alwas write our inequalit with on the left hand side and everthing else on the right hand side. We do this simplification at left. Our goal from here on out is to graph the purple inequalit at left, which is equivalent to what we started with. = To graph an inequalit, first graph the line that results when we replace the inequalit with an equal sign. This will be the boundar in our inequalit graph. This, we first graph the equation of the line at left as our boundar. If the inequalit has a > or < sign, we draw this line with a dotted line on the graph because the line itself is not part of the solution. If the inequalit has a or sign, we draw this line with a solid line because the line itself will be a part of the solution. (continued on net page) Page 1

13 m = b = 0 You can use an method ou like to graph to graph the line. For this problem, we choose the slope and -intercept method. We identif the slope and -intercept at left. Plot the -intercept at =0. The slope is m = =. 1 Find another point b starting at the - intercept and counting two units down and over to the right 1 unit. We count down initiall because the slope is negative. The line we drew was =. The original inequalit was <. This means that the actual graph that represents this inequalit consists of all points less than the line = not including the line itself (because of the < sign). You can interpret less than the line as all points below the line that ou initiall draw. We represent this b shading the graph below the line. Ans: See graph at left. Page 1

14 Question Answer 4) Graph the following inequalit: + Begin. = + rearranging in a more standard form... = + To graph an inequalit, first graph the line that results when we replace the inequalit with an equal sign. This will be the boundar in our inequalit graph. This, we first graph the equation of the line at left as our boundar. If the inequalit has a > or < sign, we draw this line with a dotted line on the graph because the line itself is not part of the solution. If the inequalit has a or sign, we draw this line with a solid line because the line itself will be a part of the solution. (continued on net page) Page 14

15 m = b = You can use an method ou like to graph to graph the line. For this problem, we choose the slope and -intercept method. We identif the slope and -intercept at left. Plot the -intercept at =. The slope is m = =. 1 Find another point b starting at the - intercept and counting two units up and over to the right 1 unit. The line we drew was = +. The original inequalit was +. This means that the actual graph that represents this inequalit consists of all points greater than the line = + including the line itself (because of the sign). You can interpret greater than the line as all points above the line that ou initiall draw. We represent this b shading the graph above the line. Ans: See graph at left. Page 15

16 Question Answer 5) Graph the following inequalit: >6 Begin. > 6 > + 6 (subtract from both sides) 6 < + (divide both sides b -) (flip inequalit sign) < (Simplif) = First, alwas write our inequalit with on the left hand side and everthing else on the right hand side. We do this simplification at left. Our goal from here on out is to graph the purple inequalit at left, which is equivalent to what we started with. We flip the inequalit sign because we divided both sides b a negative number. To graph an inequalit, first graph the line that results when we replace the inequalit with an equal sign. This will be the boundar in our inequalit graph. This, we first graph the equation of the line at left as our boundar. If the inequalit has a > or < sign, we draw this line with a dotted line on the graph because the line itself is not part of the solution. If the inequalit has a or sign, we draw this line with a solid line because the line itself will be a part of the solution. (continued on net page) Page 16

17 m = b = You can use an method ou like to graph to graph the line. For this problem, we choose the slope and -intercept method. We identif the slope and -intercept at left. Plot the -intercept at = -. The slope is m =. Find another point b starting at the - intercept and counting units up and over to the right units. The line we drew was =. The original inequalit was <. This means that the actual graph that represents this inequalit consists of all points less than the line = not including the line itself (because of the < sign). You can interpret less than the line as all points below the line that ou initiall draw. We represent this b shading the graph below the line. Ans: See graph at left. Page 17