Show $\lambda'(0)= 0 $ and $\lambda''(0) \ge 0$

Question

Consider $$\lambda(c)= \frac{1}{2}x(c)Px(c)-ax(c) \:\:where \:x(c) = (1-t)x_0 + tx_1$$ where $P\in {\mathbb{R^{dxd}}}$, and a,x$\in\mathbb{R^d}$. Show $\lambda'(0)= 0 $ and $\lambda''(0) \ge 0$.

I've gotten to $\bigtriangledown \lambda (c) = P[(1-t)x_0 + x_1 t][x_1-x_0] - b[x_1-x_0]$, but i am stumped where to go from here,