Find the values $a$, $b$ and $c$ for parabola through point $(1,1)$ and gradient $7$ at point $(3,3)$

Question

So I've been able to get here so far:

$$Let\space f(x) = ax^2 + bx + c$$ $$f '(x) = 2ax + b$$

$$f(1) = a + b + c = 1 $$ $$f(3) = 9a + 3b + c = 3$$

$$f '(3) = 6a + b = 7 $$

I don't understand how to find $a$, $b$ and $c$. I've seen the answers: $a = 3, b = -11, c = 9$. I just don't understand how to get there, given that there are three unknowns.

Answer 1

Hint:

We have three unknowns and three equations. Therefore, we can solve the system of linear equations: $$ \begin{cases} a+b+c=1\\ 9a+3b+c=3\\ 6a+b=7 \end{cases} $$

We find $a=3$, $b=-11$ and $c=9$.

Answer 2

Simultaneous equations: $$a + b + c = 1 $$ We want to get rid of one of the variables so that we can have 2 instead of 3 unknowns. I'll look at $a$. Multiply by $9$, to match the $a$ coefficient: $$9a + 9b + 9c = 9$$ Now you can subtract $f(2)$: $$9a + 3b + c = 3$$

$$9a + 9b + 9c - (9a + 3b + c) = 9 - 3$$ $$6b + 8c = 6 \tag 1$$

Do the same for $f(1)$ and $f '(3)$, but multiply $f(1)$ by $6$ to match the coefficient of $a$: $$6a + 6b + 6c = 6$$ $$6a + b = 7 $$ Subtracting: $$6a + 6b + 6c - (6a + b) = 6-7$$ $$5b+6c=-1 \tag 2$$

Now you have 2 equations with 2 unknowns: $6b + 8c = 6$ and $5b+6c=-1$
Solve these for $b$ and $c$ and then you can solve for $a$.

Answer 3

-$\begin{align}{9a+3b+c =3\\\underline{( a+b+c)=\;\;1} } \\\underline{8a+2b=\quad 2}\end{align}$

$8a+2b=2\quad 4a+b=1\\ 6a+b-(4a+b)=7-1\\ 2a=6\\a=3\\ \mathrm{And\; so \;on}$