What does $\mathbf{w^Tx}+w_0$ graphically mean in the discriminant function?
Indeed, there is a lot to understand from the $g(\mathbf{x}) = \mathbf{w}^T \mathbf{x}+w_{0}$.
So let us break down what can be geometrically interpreted:
The set $\{\mathbf{x} \mid g(\mathbf{x}) = 0\}$ is a hyperplane, if $\mathbf{w} \neq \mathbf{0}$. Then, the dimension of the hyperplane is $n-1$, if $\mathbf{x}  \in \mathbb{R}^{n}$. So for $n=2$ a hyperplane is a line, for $n=3$, the hyperplane is a plane.
The vector $\mathbf{w}$ is the normal vector. It is orthogonal to the hyperplane.
So if you take vectors $\mathbf{x}',\mathbf{x}''$ of  the hyperplane, the vector $\mathbf{x}''-\mathbf{x}'$ will be orthogonal to $\mathbf{w}$.
Thus in your figure, $\mathbf{w}$ defines the orientation of the hyperplane.
Now $w_{0}$ translates the hyperplane by a corresponding vector $\mathbf{t}$ ( see here for more details).
In the following, lets assume $||\mathbf{w}|| = 1$, which does not change the hyperplane.
We can always find a basis $\mathbf{w},\mathbf{y}_{2},\ldots,\mathbf{y}_{n}$ of $\mathbb{R}^{n}$ such that $\mathbf{w} \bot \mathbf{y}_{i}$ for all $i$, that is $\mathbf{w}$ is orthogonal to all $\mathbf{y}_{i}$.
Now given $\mathbf{x}$, there are coefficients $\lambda_{1},\ldots,\lambda_{n} \in \mathbb{R}$ such that
$\mathbf{x} = \lambda_{1} \mathbf{w} + \lambda_{2} \mathbf{y_{2}}+ \ldots + \lambda_{n} \mathbf{y_{n}}$.
Therefore $ g(\mathbf{x})-w_{0} = \mathbf{x}^{T} \mathbf{w} = \langle \mathbf{x},\mathbf{w} \rangle = \langle \lambda_{1} w+ \sum_{i = 2}^{n}\lambda_{i} \mathbf{y}_{i},\mathbf{w} \rangle = \lambda_{1} \langle w,w\rangle = \lambda_{1}$
and thus $ g(\mathbf{x}) = \lambda_{1}+w_{0}$.
The geometric conclusion is therefore: $g$ depends on how you move $\mathbf{x}$ in direction of $\mathbf{w}$.
(1) Given a vector $\mathbf{x}$, and you consider $\mathbf{x}+\mathbf{u}$ for some $\mathbf{u} \bot \mathbf{w}$, then $g(\mathbf{x}) = g(\mathbf{x}+\mathbf{u})$.
So if you move $\mathbf{x}$ &quot;along&quot; the direction of the hyperplane, $g$ does not change its value. So given $x',x''$ of the hyperplane, if you move $\mathbf{x}$ by $\mathbf{x}'-\mathbf{x}''$, you will have $g(\mathbf{x}+(\mathbf{x}'-\mathbf{x}'')) = g(\mathbf{x})$.
(2) If you move $\mathbf{x}$ further away from the plane (along the line that contains $\mathbf{w}$), the value $g(\mathbf{x})$ will either increase or decrease, depending in which direction you move. So if you move $\mathbf{x}$ by $\mu \mathbf{w}$, you will have $\mu+\lambda_{1}+w_{0} = g(\mathbf{x}+\mu \mathbf{w}) \neq g(\mathbf{x}) = \lambda_{1}+w_{0}$, if $\mu \neq 0$.
(3) $g(\mathbf{x}) = 0$ means $\lambda_{1} = -w_{0}$.
In your image, you therefore have:
(1) A point $\mathbf{x}$ on the plane has value $g(\mathbf{x}) = 0$.
(2) Each point $\mathbf{x}$ above the plane has value $g(\mathbf{x}) &gt; 0$.
(3) Each point $\mathbf{x}$ below the plane has value $g(\mathbf{x}) &lt; 0$.