## anonymous one year ago Please help Find a second degree polynomial function, f(x)=ax^2+bx+c where such that f(2)=-8, f'(2)=3, and f''(2)=4

1. jim_thompson5910

hints: f(x) = ax^2 + bx + c f ' (x) = 2ax + b f '' (x) = 2a

2. anonymous

The a, b, and c are confusing me

3. jim_thompson5910

Do you see how I got f ' (x) = 2ax + b and f '' (x) = 2a ??

$f(x)=ax^2+bx+c\\ f'(x)=2ax+b\\ f''(x)=2a$ Given $f(2) = 8=4a+2b+c \\ f'(x) =3=4a+b\\ f''(x) =4=2a \Rightarrow a=2\\$ Solving above equation, we can say: $a=2\\ 3 = 4\times 2+b \Rightarrow b=3-8=-5 \\ 4 \times 2+2\times (-5)+c=8 \Rightarrow c = 10$ Therefore $a =2 \\ b=-5 \\ c= 10$

5. anonymous

Hi @ospreytriple , what do you mean by no direct answers?

7. anonymous

Hi @sgadi . I noticed that you solved the problem for the asker. OpenStudy's code of conduct states that helpers are not to provide answers to askers but to assist and guide them so that they may be able to solve the problems for themselves.

8. anonymous

Okay so where did you get the 4 from in f(2)=-8?

@ospreytriple , I understand now. I am sorry. Shall I delete the post?

10. anonymous

Hey, I'm not a moderator. Just a heads up for future reference.