Question

f(x) = ax³ + 2x² + bx + c. Given that f(0) = f, f(1) = 4 and f(-2) = 8, find the value of a, b and c. ​

Answer 1

a = -1/6

b = 7/6

c = 1

Explanation:

`f(x)=ax^3+2x^2+bx+c\\\\f(0)=a(0)^3+2(0)^2+b(0)+c\\\\1=c`

`f(1)=a(1)^3+2(1)^2+b(1)+1\\\\4=a+2+b+1\\\\4=a+b+3\\\\1=a+b\\\\1-b=a`

`f(-2)=(1-b)(-2)^3+2(-2)^2+b(-2)+1\\\\8=-8+8b+8-2b+1\\\\8=6b+1\\\\7=6b\\\\(7)/(6)=b`

`1-b=a\\\\1-(7)/(6)=a\\ \\-(1)/(6)=a`

So, the true equation would be
`f(x)=-(1)/(6)x^3+2x^2+(7)/(6)x+1` if you substitute the coefficients and the constant.