For #f(m)=3m^2-5#, how do you find f(4),f(-3),f(6)?

Question

Nathan Axel · Accepted Answer

undefined To find $ f(4) $, $ f(-3) $, and $ f(6) $ for the function $ f(m) = 3m^2 - 5 $, you simply substitute the given values of $ m $ into the function:

1. $ f(4) = 3(4)^2 - 5 = 3(16) - 5 = 48 - 5 = 43 $
2. $ f(-3) = 3(-3)^2 - 5 = 3(9) - 5 = 27 - 5 = 22 $
3. $ f(6) = 3(6)^2 - 5 = 3(36) - 5 = 108 - 5 = 103 $

HIX Tutor · Answer

undefined To find f(4), f(-3), and f(6):
Step 1: Substitute 4 into the function f(m) = 3m^2 - 5: f(4) = 3(4)^2 - 5.
Step 2: Compute the value of f(4): f(4) = 3(16) - 5 = 48 - 5 = 43.
Step 3: Substitute -3 into the function f(m) = 3m^2 - 5: f(-3) = 3(-3)^2 - 5.
Step 4: Calculate the value of f(-3): f(-3) = 3(9) - 5 = 27 - 5 = 22.
Step 5: Substitute 6 into the function f(m) = 3m^2 - 5: f(6) = 3(6)^2 - 5.
Step 6: Determine the value of f(6): f(6) = 3(36) - 5 = 108 - 5 = 103.
Therefore, f(4) = 43, f(-3) = 22, and f(6) = 103.

Hazel Anglin · Answer

#f(4) = 43#
#f(-3) = 22#
#f(6) = 102# Given #f(color(red)(m))=3color(red)(m)^2-5# To evaluate #f(color(red)()4)# rewrite the given equation replacing each #color(red)(m)# with #color(red)(4)# and perform the arithmetic. #color(white)("XXXX")##f(color(red)(4)) = 3(color(red)(4))^2 -5# #color(white)("XXXX")##color(white)("XXXX")##=3(16)-5# #color(white)("XXXX")##color(white)("XXXX")##= 43# Similarly for #f(color(red)(-3))# #color(white)("XXXX")##f(color(red)(-3)) = 3(color(red)(-3))^2-5# #color(white)("XXXX")##color(white)("XXXX")##=3(9)-5# #color(white)("XXXX")##color(white)("XXXX")##= 22# and for #f(6)# #color(white)("XXXX")##f(color(red)(6)) = 3(color(red)(6))^2-5# #color(white)("XXXX")##color(white)("XXXX")##=3(36)-6# #color(white)("XXXX")##color(white)("XXXX")##= 102#