Evaluate $a$ and $b$ s.t. the piecewise function is differentiable at two points Evaluate $a$ and $b$ s.t. the piecewise function is differentiable at $x=1$ and $x=2$ at the same time.
\begin{equation*}
f(x) = \left\{
        \begin{array}{ll}
            ax^2+bx+1 & \quad x ≥ 1 \\
            -1 & \quad x < 1
        \end{array}
    \right.
\end{equation*}
So first, we gotta prove it's continuous at these points.
$$f(1)=a+b+1$$
$$\lim_{x \to 1^-} -1=-1$$
$$\lim_{x \to 1^+} ax^2+bx+1=a+b+1$$
$-1=a+b+1 ⟹ a+b=-2$
How do I continue this for $x=2$? And then for differentbility? Im kinda new to this and need a little help.

In order to be differentiable at $x = 1$, then the following limit needs to exits:
$$\lim_{h\to 0}\frac{f(1+h) - f(1)}{h}.$$  For this to exist you need to consider both the left and right limits:
$$\lim_{h\to 0^{-}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{-}}\frac{-1 - (a+b+1)}{h} = \lim_{h\to 0^{-}}\frac{-(a+b+2)}{h} = \lim_{h\to 0^{-}}\frac{0}{h} = 0$$ using the fact that $a + b = -2$.  Now for the right:
\begin{align*}
\lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} &= \lim_{h\to 0^{+}}\frac{a(1+h)^2 + b(1+h) + 1 - (a+b+1)}{h}\\
&=\lim_{h\to 0^{+}}\frac{a + 2ah + ah^2+b+bh+1-a-b-1}{h}\\
&=\lim_{h\to 0^{+}}\frac{2ah + ah^2 + bh}{h}\\
&=\lim_{h\to 0^{+}}2a+ah+b\\
&=2a+b.
\end{align*}
Now, for the limit to exist, you must have $2a+b = 0$, which gives you two equations in two unknowns to solve for $a$ and $b$, and I trust you can take it from here.
