How do you graph # y = (-2/3)x + 4# by plotting points?

Question

Hazel Anglin · Accepted Answer

Now draw a line passing through the points:(0, 4) and (6, 0).

#y=(-2/3)x+4# => equation of a line in slope-intercept form To draw a straight line all you need is two points on the line, then you can draw a line passing through the points. Give some values to x then solve for y: #x = 0# #y=(-2/3)*0+4# #y=4# => so point (0, 4) is on the line #x = 6# #y=(-2/3)*6+4# #y=-4+4# #y=0# => so point (6, 0) is on the line Now draw a line passing through the points:(0, 4) and (6, 0).

Genesis Allen · Answer

undefined To graph $ y = (-2/3)x + 4 $ by plotting points, you can choose several values for $ x $, plug them into the equation to find the corresponding $ y $ values, and then plot those points on a coordinate plane. For example, you could choose $ x = 0 $, $ x = 3 $, and $ x = -3 $.

When $ x = 0 $, $ y = (-2/3)(0) + 4 = 4 $, so the point is (0, 4). 
When $ x = 3 $, $ y = (-2/3)(3) + 4 = -2 + 4 = 2 $, so the point is (3, 2). 
When $ x = -3 $, $ y = (-2/3)(-3) + 4 = 2 + 4 = 6 $, so the point is (-3, 6).

Plot these points on a coordinate plane and draw a straight line passing through them.