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!

 

[daon]

So, you can start with the knowledge that it is a second degree, so the second derivative is linear and constant. Therefore, P''(2) = 2 can tell you that P''(x) = 2.

P'(x) will then be P'(x)=2x+(c1)

P(x) will be x^2 +(c1)x + (c2)

Use your initial values to solve from here...

 

[soroban]

Hello, nikchic5!

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

Obviously: \(\displaystyle \,P(x)\;=\;ax^2\,+\,bx\,+\,c\) . . . right?
We are told that: \(\displaystyle \,P(2)\,=\,5\)
\(\displaystyle \;\;\)So we have: \(\displaystyle \,a(2^2)\,+\,b(2)\,+\,c\:=\:5\;\;\Rightarrow\;\;4a\,+\,2b\,+\,c\;=\;5\)

We have: \(\displaystyle \,P'(x)\:=\:2ax\,+\,b\)
We are told that \(\displaystyle \,P'(2)\,=\,3\)
\(\displaystyle \;\;\)So we have: \(\displaystyle \,2a(2)\,+\,b\;=\;3\;\;\Rightarrow\;\;4a\,+\,b\;=\;3\)

We have: \(\displaystyle \,P''(x)\:=\:2a\)
We are told that \(\displaystyle \,P''(2)\,=\,2\)
\(\displaystyle \;\;\)So we have: \(\displaystyle \,2a\:=\:2\)

Can you solve this system? \(\displaystyle \;\begin{Bmatrix}4a\,+\,2b\,+\,c\:=\:5\\ 4a\,+\,b\:=\:3\\ 2a\:=\:2\end{Bmatrix}\)

 

[nikchic5]

Am I right?

Yea and I have that a=1 b= -1 and c=3
Is that correct? But how do I write out the polynomial? Thank so much