Jacob drove from town a to town b at an average rate of x miles per hour, then returned along the same route at y miles per hour. if he then drove back to town b at z miles per hour along the same route, what was jacob's average rate of speed for the entire trip, in miles per hour?

Average speed = (total distance traveled)/(total travel time)
= (total distance)/(time of 1st journey + time of 2nd journey + time of 3rd journey)

Let d = the distance between Town A and Town B
So, total distance traveled = 3d

Time = distance/speed
time of 1st journey = d/x 
time of 2nd journey = d/y
time of 3rd journey = d/z

Total time = d/x + d/y + dz
To simplify, rewrite with common denominator: dyz/xyz + dxz/xyz + dxy/xyz
So, total time = (dyz + dxz + dxy)/xyz

Average speed = (total distance)/(total time)
= 3d/[(dyz + dxz + dxy)/xyz]
= (3dxyz)/(dyz + dxz + dxy)
Divide top and bottom by d to get: (3xyz)/(yz + xz + xy)

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!