Second Degree.

[nikchic5]

Find the second-degree polynomial P such that P(2)=5, P'(2)=3 and P''(2)=2.

Thanks so much for anyone who can help me!S

I already have 4a+2b+c=5
4a+b=3
2a+2

Where do i go from there? Thanks so much!

 

[soroban]

Hello, nikchic5!

Find the second-degree polynomial P such that P(2)=5, P'(2)=3 and P''(2)=2.

I already have: $\displaystyle \:\begin{array}{ccc}4a\,+\,2b\,+\,c\:=\:5 \\ 4a\,+\,b\:=\:3 \\ 2a\:=\:2\end{array}$

Where do i go from there?

You've done the hard part.
You said: $\displaystyle \,P(x)\:=\:ax^2\,+\,bx\,+\,c$
You took the derivatives and plugged in $\displaystyle x\,=\,2$ and got those equations.

Now solve them.

The last one is: $\displaystyle \,2a\,=\,2\;\;\Rightarrow\;\;a\,=\,1$

Plug this into the second equation: $\displaystyle \,4(1)\,+\,b\:=\:3\;\;\Rightarrow\;\;b\,=\,-1$

Plug them into the first equation: $\displaystyle \,4(1)\,+\,2(-1)\,+\,c\:=\:5\;\;\Rightarrow\;\;c\,=\,3$

There! . . . The quadratric is: $\displaystyle \,P(x)\;=\;x^2\,-\,x\,+\,3$