Linear equations and graphs #12 - Questions and Answers

Question 1 Fill in the blank/s: When solving a system of linear equations by graphing, the system's solution is determined by locating __________. Answer Step 1 For solving a system of linear equations by graphing requires the point of intersection. When solving a system of linear equations by graphing, the system's solution is determined by locating the intersection point. Step 2 Therefore, when solving a system of linear equations by graphing, the system's solution is determined by locating Intersection point. Question 2 Explain how to graph the line 𝑥 − 𝑦 = 2 x−y=2. Answer 1 We have 𝑥 − 𝑦 = 2 x−y=2: 𝑥 − 𝑦 + 𝑦 = 2 + 𝑦 x−y+y=2+y 𝑥 = 2 + 𝑦 x=2+y 𝑥 − 2 = 0 + 𝑦 x−2=0+y 𝑥 − 2 = 𝑦 x−2=y So 𝑦 = 𝑥 − 2 y=x−2. Now, select values for 𝑥 x at random to build the graph: Let 𝑥 = − 1 x=−1, then 𝑦 = − 1 − 2 = − 3 y=−1−2=−3. Let 𝑥 = 0 x=0, then 𝑦 = 0 − 2 = − 2 y=0−2=−2. Let 𝑥 = 2 x=2, then 𝑦 = 2 − 2 = 0 y=2−2=0. So, we have the points: ( − 1 , − 3 ) , ( 0 , − 2 ) , ( 2 , 0 ) (−1,−3),(0,−2),(2,0). Now mark these points on the graph and draw a line through them. Here you are. Answer 2 We have 𝑥 − 𝑦 = 2 x−y=2. Rearrange it into standard form: 𝑦 = 𝑥 − 2 y=x−2 Select random values for 𝑥 x to make the data for your future graph: If 𝑥 = − 2 x=−2, then 𝑦 = − 2 − 2 = − 4 y=−2−2=−4. If 𝑥 = 0 x=0, then 𝑦 = 0 − 2 = − 2 y=0−2=−2. If 𝑥 = 3 x=3, then 𝑦 = 3 − 2 = 1 y=3−2=1. The points are ( − 2 , − 4 ) , ( 0 , − 2 ) , ( 3 , 1 ) (−2,−4),(0,−2),(3,1). Make a graph using these points. Question 3 What is the general form of a linear equation in two variables? Answer 1 The general form of a linear equation with two variables 𝑥 x and 𝑦 y: 𝐴 𝑥 + 𝐵 𝑦 + 𝐶 = 0 Ax+By+C=0 But, 𝐴 ≠ 0 A  =0 and 𝐵 ≠ 0 B  =0. Answer 2 𝐴 𝑥 + 𝐵 𝑦 + 𝐶 = 0 ,  when  𝐴 ≠ 0 , 𝐵 ≠ 0 Ax+By+C=0, when A  =0,B  =0 Question 4 What is the graph of 𝑓 ( 𝑥 ) = 𝑥 2 f(x)=x 2 ? Answer The graph of 𝑓 ( 𝑥 ) = 𝑥 2 f(x)=x 2 is called a "Parabola." It looks like this: (Image of a parabola provided.) One of the ways to graph this is to use plug in a few 𝑥 x-values and get an idea of the shape. Since the 𝑥 x-values keep getting squared, there is an exponential increase on either side of the y-axis. You can see this by plugging in a few values: When  𝑥 = 0 , 𝑓 ( 𝑥 ) = 0. When x=0,f(x)=0. 𝑥 = 1 , 𝑓 ( 𝑥 ) = 1 2 = 1. x=1,f(x)=1 2 =1. 𝑥 = 2 , 𝑓 ( 𝑥 ) = 2 2 = 4. x=2,f(x)=2 2 =4. 𝑥 = 3 , 𝑓 ( 𝑥 ) = 3 2 = 9. x=3,f(x)=3 2 =9. The same holds true for negative 𝑥 x-values to the left of the y-axis since a negative value squared is positive. For example: 𝑥 = − 1 , 𝑓 ( 𝑥 ) = ( − 1 ) 2 = 1. x=−1,f(x)=(−1) 2 =1. 𝑥 = − 2 , 𝑓 ( 𝑥 ) = ( − 2 ) 2 = 4. x=−2,f(x)=(−2) 2 =4. Question 5 How can I graph 𝑦 = 2 𝑥 − 1 y=2x−1 by plotting points? Answer 1 To find the y-intercept, put 𝑥 = 0 x=0: 𝑦 = 2 ( 0 ) − 1 y=2(0)−1 𝑦 = − 1 y=−1 To find the x-intercept, put 𝑦 = 0 y=0: 2 𝑥 − 1 = 0 2x−1=0 2 𝑥 = 1 2x=1 𝑥 = 1 2 x= 2 1 ​ Draw a system of axes. Mark these points on the graph. Now draw a line through these two points. To draw the graph by used points, Let 𝑥 = − 1 x=−1, then 𝑦 = 2 ( − 1 ) − 1 = − 2 − 1 = − 3 y=2(−1)−1=−2−1=−3. Thus, first point is ( − 1 , − 3 ) (−1,−3). Let 𝑥 = 0 , 𝑦 = − 1 x=0,y=−1. And second point is ( 0 , − 1 ) (0,−1). Let 𝑥 = 1 , 𝑦 = 2 ( 1 ) − 1 = 1 x=1,y=2(1)−1=1. Third point is ( 1 , 1 ) (1,1). Answer 2 When the exponents of the variables are all 1, the equation is a line. Thus, it is enough to find two points to draw it. We can calculate two points assigning a number to 𝑥 x and calculating the corresponding number for 𝑦 y. For example, when 𝑥 = 0 , 𝑦 = 2 ⋅ 0 − 1 = − 1 x=0,y=2⋅0−1=−1. And the line will pass through the point ( 0 , − 1 ) (0,−1). Now, we need a second point, we can take 𝑥 = 1 x=1, for example, then 𝑦 = 2 ⋅ 1 − 1 = 1 y=2⋅1−1=1. The line passes through ( 1 , 1 ) (1,1). Finally, we can mark the two points on the plane and draw the line passing from these points.