Given function $f: \mathbb{R^n} \to \mathbb{R}$ is second-order differentiable and a constant $L>0$. Prove that the following formula is quadratic approximation of $f$ at point $y$ $$f(y) + \langle \nabla f(y), x-y\rangle + \frac{L}{2}\Vert x-y \Vert^2.$$
My attempt: Let $g(t) = f(tx + (1-t)y),\ t \in [0,1]$. Using Taylor approximation with Langrange residuals for $g$ at $t=0$ we have $$g(1) = g(0) + g'(0) + \frac{1}{2}g''(\theta),$$ where $\theta \in (0,1)$. Applying the chain rule we have $$f(x) = f(y) + \langle\nabla f(y),x-y \rangle + $$
I'm stuck here since I don't know how to calculate $g''(\theta)$. May anyone can help me?