Determining a function given points and slopes

Question

A polynomial function of degree 2 passes through the point P(1, 3). At this point the function’s slope (i.e. the slope of its tangent) is 1 and at the point Q(2, y) it has a slope of 5. Determine the function.

I am not sure how to approach this question. I have found the equation for the tangent lines: y = x + 2 and y = 5x - 10 + y for points P and Q respectively. I know that the function will have the general form $(x) = ax^2 + bx + c$. However, I am not sure how to implement this information in order to find the function. I know I am supposed to use derivatives but not quite sure how.

Answer 1

HINT

Let $P(x) = ax^{2} + bx + c$. According to the given data, we can conclude that

\begin{align*} \begin{cases} P(1) = a + b + c = 3\\\\ P'(1) = 2a + b = 1\\\\ P'(2) = 4a + b = 5 \end{cases} \end{align*}

Can you take it from here?

Answer 2

Let $f(x)=ax^2+bx+c$, then $a+b+c=3$. $f'(x)=2ax+b$, then $2a+b=1, 4a+b=5 \implies a=2, b=-3$, then $c=4$. So $f(x)=2x^2-3x+4$.

Answer 3

$$ F(x)=ax^2+bx+c\\ F'(x)=2ax+b $$ At $(1,3)$ we have $F(1,3)=a+b+c=3$ and $F'(1)=2a+b=1$. At $(2,y)$ we have $F'(2)=4a+b=5$ Thus we have 3 equations with 3 unknowns that we can solve to find: $$ a = 2,\ \ b = -3,\ \ c = 4 $$