UNIT 3B CREATING AND GRAPHING EQUATIONS Lesson 4: Solving Systems of Equations Instruction

Transcription

1 Prerequisite Skills This lesson requires the use of the following skills: graphing multiple equations on a graphing calculator graphing quadratic equations graphing linear equations Introduction A system of equations is a set of equations with the same unknowns. A quadratic-linear system is a system of equations in which one equation is quadratic and one is linear. A quadratic equation can be a parabola or a circle. We learned previously how to solve systems of linear equations. In this lesson, we will learn to solve a quadratic-linear system by graphing. Key Concepts A circle is the set of all points that are equidistant from a reference point, the center. The set of points forms a two-dimensional curve that measures 360. A circle is another form of a quadratic equation. The standard form of an equation of a circle is (x h) + (y k) = r, where (h, k) is the center and r is the radius. If the circle is centered at (0, 0), then the equation of the circle has the form x + y = r. A quadratic-linear system may have one real solution, two real solutions, or no real solutions. The solutions of a quadratic-linear system can be found by graphing. The points of intersection of the graphed quadratic and linear equation are the ordered pair(s) where graphed functions intersect on a coordinate plane. These are also the solutions to the system. U3B-197

2 Graphed Quadratic-Linear Systems Two real solutions One real solution No real solutions One real solution No real solution Real-world problems may appear to have two solutions when graphed, but in fact only have one because of the context of the problem. Traditionally, negative values are ignored. For example, if graphing the distance a car traveled in relation to time, if one algebraic solution had a negative value, it would not be a true solution since distance and time are always positive. Common Errors/Misconceptions not identifying all points of intersection incorrectly entering the equations into the calculator U3B-198

3 Guided Practice Example 1 Graph the system of equations below to determine the real solution(s), if any exist. y= x + x y= x 1. Graph the quadratic function, y = x + x. If graphing by hand, first plot the y-intercept, (0, ). Next, use the equation for the axis of symmetry to determine the x-coordinate of the vertex of the parabola. b x = = = = 1 a 1 ( ) Draw the axis of symmetry, x = 1. Substitute the x-coordinate to determine the corresponding y-coordinate of the vertex. y = ( 1) + ( 1) = 1 = 3 Plot the vertex, ( 1, 3). Finally, since the y-intercept is 1 unit right of the axis of symmetry, an additional point on the parabola is 1 unit left of the axis of symmetry at (, ). Sketch the curve. If using a graphing calculator, follow the directions appropriate to your model. On a TI-83/8: Step 1: Press [Y=]. Step : At Y 1, type [X, T, θ, n][x ][+][][X, T, θ, n][ ][]. Step 3: Press [GRAPH]. On a TI-Nspire: Step 1: Press [menu] and select 3: Graph Type, then select 1: Function. (continued) U3B-199

4 Step : At f1(x), enter [x], hit the [x ] key, then type [+][][x][ ][] and press [enter] to graph the quadratic function y x 6 8. Graph the linear function, y = x, on the same grid. If graphing by hand, first plot the y-intercept, (0, ). Note: Since the two functions have the same y-intercept, we have already determined one solution. Next, use the slope of to graph additional points in either direction. Sketch the line. On a TI-83/8: Step 1: Press [Y=]. Step : At Y, type [][X, T, θ, n][ ][]. Step 3: Press [GRAPH]. On a TI-Nspire: Step 1: Press [menu] and select 3: Graph Type, then select 1: Function. (continued) U3B-00

5 Step : At f (x), enter [][x][ ][] and press [enter] to graph the linear function y (0, ) x Note any intersections that exist as the solution or solutions. On a TI-83/8: Step 1: To approximate the intersections, press [nd][trace] and select intersect. Step : Press [ENTER]. Step 3: At the prompt that reads First curve?, press [ENTER]. Step : At the prompt that reads Second curve?, press [ENTER]. Step 5: At the prompt that reads Guess?, arrow as close as possible to the first point of intersection and press [ENTER]. Write down the approximate ordered pair. On a TI-Nspire: Step 1: Press [menu] and select 7: Points & Lines, then select 3: Intersection Point(s). Step : Select the graphs of both functions. The only point of intersection is (0, ). y= x + x The solution of is (0, ). y= x U3B-01

6 Example Graph the system of equations below to determine the real solution(s), if any exist. y= 3x y= x + x 1. Graph the quadratic function, y = x + x, using the methods demonstrated in Example y x 6 8 U3B-0

7 . Graph the linear function, y = 3x, on the same grid y 6 (1, 3) x (, 6) Note any intersections that exist as the solution or solutions. The two points of intersection are (, 6) and (1, 3). y= 3x The solutions of are (, 6) and (1, 3). y= x + x U3B-03

8 Example 3 Graph the system of equations below to determine the real solution(s), if any exist. y= x 10 y= x x 6 1. Graph the quadratic function, y = x x 6, using the methods demonstrated in Example 1. y x U3B-0

9 . Graph the linear function, y = x 10, on the same grid. y x Note any intersections that exist as the solution or solutions. The two functions do not intersect. y= x 10 The system has no real solutions. y= x x 6 U3B-05

10 Example Solve the system that appears to be represented by the circle shown and the line with the given equation. 3 y= x Graph the line on the same coordinate plane as the circle. To graph the line by hand, first graph the y-intercept, (0, 9). Then use the slope to plot the next point. The equation is already written in slope-intercept form, y = mx + b, so 3 m =. Because the slope is positive, it can be rewritten with two negatives since a negative divided by a negative is a positive. 3 3 m = = (continued) U3B-06

11 Starting from the y-intercept, plot the next point by moving down 3 units and to the left units. The next point is located at (, 6). From (, 6), plot the next point by again moving down 3 units and to the left units to plot point (, 3). From (, 3), plot the next point by moving down 3 units and to the left units to plot the point ( 6, 0). Use a straightedge to connect the points. ( 6, 0) (, 3) (, 6) (0, 9) Determine the point(s) of intersection, if any. The graph of the line appears to intersect the circle at one point, (, 3). Therefore, the solution to the system represented by the graphed 3 circle and the linear equation y= x +9 is (, 3). U3B-07

12 Name: UNIT 3B CREATING AND GRAPHING EQUATIONS Date: Problem-Based Task : Constant Acceleration and Constant Speed A car begins at rest and accelerates. Its distance from its starting position in meters, C(t), can be determined as a function of time in seconds, t, by the formula C(t) = t. A second car, 00 meters ahead, is traveling at a constant speed of 0 meters per second. Its distance from the first car s starting point, C (t), in meters can be determined as a function of time, t, in seconds by the formula C (t) = 0t How long after the first car accelerates will the cars be side by side? How long after the first car accelerates will the cars be side by side? U3B-08

13 Name: UNIT 3B CREATING AND GRAPHING EQUATIONS Date: Problem-Based Task : Constant Acceleration and Constant Speed Coaching a. What are the unknowns? b. What do you need to know in order to answer the question posed in the task? c. How will you determine the solution? d. How long after the first car accelerates will the cars be side by side? U3B-09

14 Problem-Based Task : Constant Acceleration and Constant Speed Coaching Sample Responses a. What are the unknowns? Time and distance are the unknowns. b. What do you need to know in order to answer the question posed in the task? You need to know the time, t, after the first car accelerates until both cars are the same distance from the first car s starting point, C(t). c. How will you determine the solution? Graph the functions and find the point(s) of intersection. d. How long after the first car accelerates will the cars be side by side? The following graph shows the points of intersection. 500 C(t) 00 Distance (meters) Time (seconds) t There are two points of intersection, but because the quantities are distance and time, disregard the negative values. According to the graph, the cars will be side by side about 16. seconds after the first car accelerates. Recommended Closure Activity Select one or more of the essential questions for a class discussion or as a journal entry prompt. U3B-10