34. As with many problems, substitute the given information into the given formula and solve for the variable of interest. Here, it can be helpful to rewrite f(x) as a square.
... f(x) = x² +4x +4 = (x +2)²
Substituting in accordance with the problem statement, you have
... f(x-h) = x² = (x -h +2)²
In order to make the term on the right match the term in the middle, we must choose h=2. This gives
... f(x -2) = x² = (x -2 +2)²
_____
You can also work it out the long way.
... f(x -h) = (x -h)² +4(x -h) +4 = x²
... x² -2hx +h² +4x -4h +4 = x² . . eliminate parentheses
... x(4 -2h) +(4 -4h +h²) = 0 . . . . . subtract x² and collect terms
We know both the coefficient of x and the constant must be zero, so we have two equations:
... 4 -2h = 0 ⇒ h = 2
... h² -4h +4 = 0 = (h -2)² ⇒ h = 2
35. Use the formula for the volume of a cylinder, fill in the given information, and solve for the unknown value. If you use a formula that involves radius, you must then convert that to diameter. (Diameter is twice the radius.)
The volume is given by
... V = π·r²·h
Since the can is half full, h = (1/2)·(4 in) = 2 in. Substituting the given values, we have
... 32π in³ = π·(2 in)·r²
... 16 in² = r² . . . . . divide by 2π in
... √(16 in²) = r = 4 in . . . . take the square root. This gives the radius.
The diameter of the can is 8 inches,
