Recovering a quadratic polynomial from three values using calculus

I'm asked to solve this using calculus:

Let $$ f(x) = ax^2 + bx +c .$$ If $ f(1) = 3 $, $f(2) = 7$, $f(3) = 13$, then find $a$, $b$, and $f(0)$.

I know I can solve this using solving three equations simultaneously. And I can also solve this using Gauss Jordan or Gaussian elimination method by writing the augmented matrix. But I'm wondering is there any other method to solve this.

Solving by any method it turns out that $a = b = c = 1$.

This problem has nothing to do with calculus. Knowing about the symmetries of the quadratic function one can proceed as follows:

Make $(2,7)$ your origin. This amounts to introducing the function $$g(y):=f(2+y)-7\ .$$ Then $$g(y)=a'y^2+b'y+c', \qquad g(-1)=-4,\quad g(0)=0,\quad g(1)=6\ ,$$ and therefore $$c'=0,\qquad 2a'=g(1)+g(-1)=2,\qquad 2b'=g(1)-g(-1)=10\ .$$ It follows that $g(y)=y^2+5y$, so that $$f(x)=g(x-2)+7=(x-2)^2+5(x-2)+7=x^2+x+1\ .$$

Hint

We can use the following convenient geometric fact about the graphs of polynomials of degree $\leq 2$:

Lemma The slope $m$ of the secant line between points $(x_1, y_1)$ and $(x_2, y_2)$ on the graph of a polynomial function $f(x) := a x^2 + b x + c$ of degree $\leq 2$ is the slope of the tangent line to the graph of $f$ at the midpoint of the interval $[x_1, x_2]$, that is, $m = f'\left(\frac{x_1 + x_2}{2}\right)$.

Proof By definition, $$m := \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{(a x_2^2 + b x_2 + c) - (a x_1^2 + b x_1 + c)}{x_2 - x_1} = a (x_1 + x_2) + b,$$ but we can write this as $$2a\left(\frac{x_1 + x_2}{2}\right) + b = f'\left(\frac{x_1 + x_2}{2}\right) .$$ QED.

This lemma immediately gives $$f'\left(\tfrac{3}{2}\right) = f(2) - f(1), \qquad f'(2) = \frac{f(3) - f(1)}{2}, \qquad f'\left(\tfrac{5}{2}\right) = f(3) - f(2) .$$ Now, $f'$ is itself linear, so it satisfies the same property, and hence we can recover the second derivative of $f$ at a convenient point.

Computing gives $$f''(2) = f'\left(\tfrac{5}{2}\right) - f'\left(\tfrac{3}{2}\right) = [f(3) - f(2)] - [f(2) - f(1)] = f(3) - 2 f(2) + f(1) .$$ We now have $f(2), f'(2), f''(2)$, which lets us recover $f(x)$ from its Taylor polynomial at $x = 2$: $$f(2) + f'(2) (x - 2) + \tfrac{1}{2} f''(2) (x - 2)^2 .$$ Substituting and simplifying gives $f(x) = x^2 + x + 1$.