Answer with Step-by-step explanation:
We are given that:
f(x)=ax²+bx+c
also at x=0 f(x)=15
Putting x=0 in f(x)
a×0+b×0+c=15
⇒ c=15
now, f(2)=15.5
⇒ a×2²+b×2+c=15.5
⇒ 4a+2b+15=15.5
⇒ 4a+2b=0.5 -------(1)
f(4)=17
⇒ a×4²+b×4+c=17
⇒ 16a+4b+15=17
⇒ 16a+4b=2 --------(2)
equation (2)- 2×equation(1)
⇒ 16a+4b-8a-4b=2-2×0.5
⇒ 8a=2-1
⇒ 8a=1
⇒ a=1/8
⇒ a=0.125
Putting value of a and c in equation (1)
⇒ 4×0.125+2b=0.5
⇒ 0.5+2b=0.5
⇒ 2b=0
⇒ b=0
Hence
a=0.125
b=0
c=15
and f(x)=0.125x²+15