Question

find all polynomials f(t) of degree < or = to 2 whose graphs run through the points (1,1) and (3,3), such that f '(2) = 3. Use f(t) = ax^2 + bx + c and solve linear systems for a, b, c.
This question was asked before, and the solution arrived at was that there was none as two of the equations were the same. Could someone confirm please?

Answer 1

It can't be a 0th degree polynomial, because those two points require a line.
It can't be a 1st degree polynomial, because the slope be constant at: (3-1)/(3-1) = 1
The derivative says it would need a slope of 3 when x = 2.

So if there is a solution, it would have to be a 2nd degree polynomial

Answer 2

\[
\begin{array}{rcl}
f(t) &=& at^2+bt+c \\
f'(t) &=& 2at + b

\end{array}
\]

Answer 3

@skaplan1995 Why are there \(x\)?
\( f(t) = ax^2 + bx + c\)

Answer 4

\[
\begin{array}{rcl}
f'(t) &=& 2at+b \\

f'(2) &=& 2a(2)+b \\

3 &=& 4a+b \\
3-4a &=& b \\

\end{array}
\]Leaving us with: \[
f(t) = at^2+(3-4a)t+c
\]

Answer 5

\[
\begin{array}{rcl}
f(t) &=& at^2+(3-4a)t+c \\
f(1) &=& a(1)^2+(3-4a)(1)+c \\
1 &=& a+(3-4a)+c \\
1 &=& 3-3a+c \\
1- (3-3a) &=& c \\
3a-2 &=& c \\

\end{array}
\]Finally we have: \[
f(t) = at^2+(3-4a)t+(3a-2)
\]

Answer 6

@skaplan1995 What do you think?

Answer 7

This is fantastic, thank you very much!