Which of the following exponential functions go through the points (1,12) and (2,36)?

a. f(x) = 4(3)^-x
b. f(x) = 3(4)^-x
c. f(x) = 3(4)^x
d. f(x) = 4(3)^x

D

Step-by-step explanation:

Evaluate both points into each equation and see which works out..

A:  f(1) = 4(3)^-1 = 4/3, it's supposed to be 12, so this equation is ruled out

B:  f(1) = 3(4)^-1 = 3/4, ruled out for the same reason as A

C:  f(1) = 3(4)^1 = 12, this works, now check the 2nd point...

    f(2) = 3(4)² = 3(16) = 48, it's supposed to be 36, so this equation is ruled out

D:  f(1) = 4(3)^1 = 12, this works, now check the 2nd point..

    f(2) = 4(3)² = 4(9) = 36, this is what we wanted.  

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)