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: \(\displaystyle \,P(x)\;=\;x^2\,-\,x\,+\,3\)
