0 If fz 2 2r 5 then fz h A2 225h B2 254M C2 2xh2 2h 5 D2

0. If f(z) 2 + 2r + 5, then f(z + h)- (A)2 + 22+5+h (B)2 +2+54M (C)2 +2x+h2 +2h +5 (D)2 +Zr+2zh + 2h + h2 + 5 (E) T2 +22+2h + 7h + h2 +5

Solution

f(x) = x^2 + 2x + 5

f(x + h) = (x + h)^2 + 2(x + h) + 5

Now, we have to simply expand that...

(x + h)^2 + 2(x + h) + 5

(x+h)^2 :
This is (x + h)(x + h), which we then FOIL to get
x^2 + xh + xh +h^2
x^2 + 2xh + h^2

2(x+h) :
This expands as 2x + 2h

So, problem now becomes :
x^2 + 2xh + h^2 + 2x + 2h + 5

So, option D