Can someone help me evaluate functions $f(x) \in \mathbb{R}[x]$ at the points in $\text{Spec}\, \mathbb{R}[x]$ ?
$(x-a) \in \text{Spec} \,\mathbb{R}[x]$ so I think we still have a point there...
$(x^2 +ax+b) \in \text{Spec} \,\mathbb{R}[x]$ is a maximal ideal as well (if $a^2 < 4b$) and therefore also a point.
In the first case, we can value the function using the Euclidean algorithm:
$$ f(x) \equiv \;\;\;\;f(a) + (x-a)\,\mathbb{R}[x]$$
but what about the other points. Can we evaluate $f(x)$ "at" the other maximal ideals such as $(x^2 + ax + b)$?
$$ f(x) \equiv \;? + (x^2 + ax + b)\,\mathbb{R}[x]$$
Well, you can use polynomial long division to compute the remainder of $f(x)$ when dividing by $x^2+ax+b$. The remainder will be some linear polynomial, and is a canonical representative of the residue class of $f(x)$ mod $(x^2+ax+b)$.
You can get a more concrete picture of this by identifying the residue field with $\mathbb{C}$. Namely, if $\alpha\in\mathbb{C}$ is either of the two complex roots of $x^2+ax+b$, then we get an isomorphism $\mathbb{R}[x]/(x^2+ax+b)\to\mathbb{C}$ sending $x$ to $\alpha$. This isomorphism maps the residue class of $f(x)$ to $f(\alpha)$. So, if we identify the residue field with $\mathbb{C}$, we can "evaluate" $f(x)$ at this point by just evaluating the polynomial at $x=\alpha$.