2. given the equation 2x – 3y + 4z =12, find the three intercepts for this plane and graph it below. show your work. view the video “how to submit a part 2” for on how to graph. (2 point for each intercept, 3 for the graph.)

Given the equation 2x-3y+4z=12, to calculate the intercepts we proceed as follows:
x-intercept:
To get x-intercept we let y=0 and z=0. The substitute these values in our equations follows:
2x+3(0)+4(0)=12
2x=12
hence:
x=12/2
x=6
therefore the x-intercept will be at point (6,0,0)

y-intercept:
To calculate for y-intercept we let x=0 and z=0 and substitute the values in the equation, here we shall have:
2(0)-3y+4(0)=12
solving for y we get:
-3y=12
hence:
y=12/(-3)
y=-4
therefore the y-intercept is at point (0,-4,0)

z-intercept:
To get the z-intercept we let x=0 and y=0, the we substitute the values in the equation as follows:
2(0)-3(0)+4z=12
solving for z we get:
4z=12
z=12/4
z=3
hence the z-intercept is at (0,0,3)

6(2x + 5y)6 (2x) = 12x6 (5y) = 30y12x + 30y