Point $A (9,3)$ lies on the graph of $h(x) = \sqrt x $. State the coordinates of $A'$, the image of $A$ on $g(x) = 1/2h(x+3)-4$.
First, I wrote $g(x) = 1/2 * 6 - 4 = -1$. Yet, my teacher wrote $g(x)=1/2(3)-4$. She wrote $A'$ as $(6, -2.5)$, but I cannot figure out why she wrote her equation like that, even with her explanation.
Why did she write her equation like that, and how did she find the x-coordinate for $A'$?
In the stated form, the task is ill-posed, and open to interpretation - it's not mathematically rigorous unless it's specified what exactly the "image" means. If you are mapping 2D to 2D (coordinates to new coordinates), you need more than one function. You would need to specify how both $x$ and $y$ are transformed.
However, I think I can decipher what kind of transformation she meant. The trick is not to put $h(x)=\sqrt{x}$ into it but to literally follow transformation of the function graph step by step. Consider this sequence:
$$h(x)\to h(x\color{red}{+3})$$ $$h(x+3)\to \color{red}{\frac12} h(x+3)$$ $$\frac12h(x+3)\to \frac12h(x+3)\color{red}{-4}$$
First step is shift to left by 3, then scale by half vertically and then shift by 4 down. Therefore, we get
$$(9,3)\to(6,3)\to (6,3/2)\to (6,-5/2)$$
Still, you were right to question the meaning of the task. If you just see the written function as $g(x)=\frac12 \sqrt{x+3}-4$, you can just put in a single number $x$ and get another number out (no way of knowing which original 2D point you came from). Similarly, if someone wrote $\frac12 h(x+3)-4h(x)$, you could not interpret it the way I did (you would have a combination of two shifted copies of the same curve). You should treat this just as a very effective learning task for getting used to shifting and scaling functions in all directions.