Instead of thinking about every point on the graph, just track what happens to the vertex when you move the graph left or right, up or down.

Start from the vertex of $f$: how does moving 6 units left change its $x$-coordinate, and how does moving 2 units down change its $y$-coordinate?

Once you know the new vertex $(h,k)$, plug it into $(x-h)^2+k$ and compare that expression to the answer choices.

In Desmos, enter f(x) = (x - 4)^2 + 3. Look at the graph and note the vertex that appears at $(4,3)$.

To shift $f$ 6 units left, add 6 inside the input: type y1 = f(x + 6). Check the new vertex coordinates shown on the graph.

Now shift that graph 2 units down: type y2 = f(x + 6) - 2. Observe the vertex of this final graph and, if Desmos simplifies the expression, compare the displayed formula of y2 with the answer choices to see which one matches.

You can also enter each answer choice as a separate function (for example, gA(x) = (x + 2)^2 - 1, gB(x) = (x - 2)^2 + 1, etc.) and see which graph coincides exactly with y2 = f(x + 6) - 2.

Notice that $f(x)=(x-4)^2+3$ is in vertex form:

where the vertex is $(h,k)$.

Compare $f(x)$ with this form to find its vertex.

From $f(x)=(x-4)^2+3$:

• The term $(x-4)$ means $h=4$ (remember, the sign is reversed inside the parentheses).
• The constant term $+3$ means $k=3$.

So the vertex of $f$ is $(4,3)$.

We are told the graph of $f$ is shifted:

• 6 units to the left: subtract 6 from the $x$-coordinate.
• 2 units down: subtract 2 from the $y$-coordinate.

Starting from the original vertex $(4,3)$:

• New $x$-coordinate: $4-6=-2$.
• New $y$-coordinate: $3-2=1$.

So the vertex of $g$ is $(-2,1)$.

A quadratic with vertex $(h,k)$ in vertex form is

Here $h=-2$ and $k=1$, so

Among the answer choices, this matches choice D: $g(x)=(x+2)^2+1$.