Raymond J. answered 02/17/21
Patient with Ability to Explain in Many Ways
When graphing these equations, pick some points for x then figure out what y is. Start simple, like x = 0 (this is the vertical intercept, or y-intercept, ie when x = 0 it crosses (intercepts) the vertical axis or the y-axis).
For the first equation, f(x) = (5)(1/5)x let's simply plot some points:
f(0) = 5(1/5)0 = 5
f(1) = (5)(1/5)1 = (5/5) = 1
f(-1) = (5)(1/5)-1 = (5)(5) = 25
f(2) = (5)(1/5)2 = (5/25) = 5
f(3) = (5)(1/5)3 = (5/125) = 1/25
Once the points are plotted we can get an idea of what the graph will look like. We can see that it extends sharply upward as it goes left of the y-axis, but as it goes further right along the x-axis it squashes down, but never touches the x-axis.
For the second one, I'm not sure if it's f(x) = 3x - 2 or f(x) = 3x-2
No matter which, just plot some points. Start with x = 0, then try x = 1 etc.
f(x) = 3x - 2
f(0) = 30 - 2 = 1 - 2 = -1
f(1) = 31 - 2 = 3 - 2 = 1
f(-1) = 3-1 - 2 = 1/3 - 2 = -(5/3)
f(2) = 32 - 2 = 9 - 2 = 7
Plot these points on the graph then draw out the curve.
Similarly for f(x) = 3x-2
f(0) = f(x) = 30-2 = 1/9
f(1) = f(x) = 31-2 = 1/3
f(2) = f(x) = 32-2 = 1
f(3) = f(x) = 33-2 = 3
f(4) = f(x) = 34-2 = 9
f(-1) = f(x) = 3-1-2 = 1/27