a Plot both points to graph the line. Try finding the linear equation with the given information in order to fill the table.
B
b Isolate y and compare Mica's idea to the equation found in Part A.
A
aGraph:
Table:
x
y
-3
3
- 2
1
- 1
- 1
0
-3
1
-5
2
-7
3
-9
B
b She is correct, see solution.
Practice makes perfect
a The exercise says that Mica lost almost all of the information related to the linear equation, but she still had the following parts of the graph and table.
x
y = -2x -3
-3
-2
1
-1
0
1
2
3
We are asked to complete the table and the graph. Even though most of the information was lost, we only need two points to define a linear equation. We have two points — from the graph we have (-3,3), and from the table we have the point (-2,1). We can plot the point from the table in the graph, then draw the line passing through both points.
Recall the Slope Formula.
m = y_2-y_1/x_2-x_1
Here m is the slope and (x_1,y_1) and (x_2,y_2) are two known points. We can use it together with the known points to find the slope of our function.
We can use the slope-intercept form of the line to find our linear equation.
y = mx+b
Here m is the slope and b is the y-intercept. By looking at the graph again, we can identify the y-intercept as -3.
Now that we found the slope and know the y-intercept, we can write the linear equation.
y = -2x -3
We will use it to complete the table.
x
y=-2x -3
y
- 3
y = (-2)( - 3) -3
3
- 2
y = (-2)( - 2) -3
1
- 1
y = (-2)( - 1) -3
- 1
0
y = (-2)( 0) -3
-3
1
y = (-2)( 1) -3
-5
2
y = (-2)( 2) -3
-7
3
y = (-2)( 3) -3
-9
b Mica thinks that the equation for the graph can be 2x+y = -3. We are asked to explain if she is correct or not and why. As we can see from Part A, the equation for the linear function was y = -2x -3. Let's isolate y from Mica's suggestion to see if they are equivalent.
