
William W. answered 12/11/19
Math and science made easy - learn from a retired engineer
For each of these, just plug in the "thing" in parenthesis into the function wherever there is an "x". So for f(y+3) you would plug in "y + 3" in for x like this:
f(y+3) = 4(y + 3)2 − 5(y + 3) + 2 and simplify:
f(y+3) = 4(y2 + 6y + 9) − 5y − 15 + 2
f(y+3) = 4y2 + 24y + 36 − 5y − 13
f(y+3) = 4y2 + 19y + 23
Do the same for the other problem except first plug in "1 + h" and then plug in "1" and then subtract.
f(1 + h) = 4(1 + h)2 − 5(1 + h) + 2
f(1 + h) = 4(1 + 2h + h2 − 5 − 5h + 2
f(1 + h) = 4 + 8h + 4h2 − 3 − 5h
f(1 + h) = 4h2 + 3h + 1
Then do f(1) = 4(1)2 − 5(1) + 2 = 4 - 5 + 2 = 1
The subtract to get:
f(1 + h) - f(1) = (4h2 + 3h + 1) - (1)
f(1 + h) - f(1) = 4h2 + 3h