Focus on $(x-1)^2$. What is the smallest value a squared expression can have, and for which $x$ does that happen?

Once you know the smallest value of $(x-1)^2$, substitute it into $2(x-1)^2+3$ to find the smallest possible value of $g(x)$.

In Desmos, type y = 2(x - 1)^2 + 3 to graph the parabola.

Zoom or move the graph, then tap/click on the lowest point of the parabola (its vertex). Desmos will show the coordinates of this point.

Look at the $y$-coordinate of the vertex you just found; that $y$-value is the minimum value of $g(x)$, and you should choose the answer option that matches this number.

Notice that $g(x)=2(x-1)^2+3$ is a quadratic written in vertex form:

Here, $a=2$, $h=1$, and $k=3$. This means the graph is a parabola that opens upward (because $a>0$) and has its vertex at $(h,k)=(1,3)$.

For any real number $x$, a square is never negative:

Multiplying by $2$ keeps it nonnegative:

Now add $3$ to both sides:

So $g(x) = 2(x-1)^2 + 3$ is always at least $3$ for any $x$.

The inequality $2(x-1)^2 + 3 \ge 3$ becomes an equality (i.e., $g(x)=3$) when the squared part is as small as possible, which is $0$.

Set $(x-1)^2 = 0$:

Then

This is the smallest possible value of $g(x)$, so the minimum value of $g$ is 3.