For a quadratic $ax^2+bx+c$, the x-coordinate of the vertex is given by $x=-\dfrac{b}{2a}$. Identify $a$ and $b$ in this function and compute that x-value.

Once you know the x-coordinate of the vertex, substitute that x-value back into $f(x)$ and simplify carefully. That result is the minimum value of the function.

In Desmos, type y = x^2 - 48x + 2304 to graph the function.

Zoom or pan until you can clearly see the lowest point of the parabola. Tap or click on that lowest point (the vertex); Desmos will display its coordinates $(x,y)$.

Look at the y-coordinate of the vertex shown by Desmos. That y-value is the minimum value of the function $f(x)$.

The function $f(x)=x^2-48x+2304$ is a quadratic of the form $ax^2+bx+c$ with $a=1>0$, so its graph is a parabola that opens upward.

For an upward-opening parabola, the minimum value of the function occurs at its vertex. So we need to find the vertex and then find the function value there.

For a quadratic $ax^2+bx+c$, the x-coordinate of the vertex is given by

Here, $a=1$ and $b=-48$, so

So the vertex occurs at $x=24$.

Now plug $x=24$ into $f(x)$ to find the minimum value:

Since the parabola opens upward, this value is the minimum value of the function, which is 1728.