Can someone explain how to use mapping (this method specifically please!) to figure out the sequence of transformations that takes the graph of y = 2\(\sqrt{4-x}\) + 3 to the graph of y= -\(\sqrt{x}\) + 6? Thank you!!

Edit: fixed the formatting

I'm not sure what you mean in your question about "how to use mapping" regarding transformations. Could you please elaborate? Nonetheless, I'll answer the question a few different ways.

You should note that there are often

*several* possible sequences of transformations that will end in the same outcome. Obviously some are simpler than others, but we should keep that in mind. For example, for \(y=x^2\), a dilation by factor \(4\) from the \(x\)-axis is the

*same* as a dilation by factor \(1/2\) from the \(y\)-axis.

**Method 1: Using dash notation**First, let's obtain expressions for the image of \(x\) and \(y\) (\(x'\) and \(y'\) respectively).

We can rewrite both equations to get \(\dfrac{y-3}{2}=\sqrt{4-x}\) and \(6-y=\sqrt{x}\).

Thus, we have \[\begin{cases}x'=4-x\\ 6-y'=\displaystyle\frac{y-3}{2}\end{cases}\implies\begin{cases}x'=-x+4\\ \displaystyle y'=-\frac{1}{2}y+\frac{15}{2}\end{cases}.\]Once we have it in this form, we can just read off the transformations. Remember, convention is that we give them in the order: dilations, reflections, translations.

1) Dilation by factor \(1/2\) from the \(x\)-axis

2) Reflection in the \(x\)-axis

3) Reflection in the \(y\)-axis

4) Translation of \(4\) units in the positive \(x\)-direction

5) Translation of \(15/2\) units in the positive \(y\)-direction

**Method 2: Using some clever intuition**For our convenience, let's define \(f(x)=2\sqrt{4-x}+3\) and \(g(x)=-\sqrt{x}+6\).

Let's go through dilations. Clearly, there is an unwanted \(2\) in the rule of \(f\) in front of the square root. To get rid of that, let's apply a dilation by factor \(1/2\) from the \(x\)-axis. That is, after the first transformation, the rule of \(f\) becomes \[f_1(x)=\frac12 f(x)=\frac{1}{2}\big(2\sqrt{4-x}+3\big)=\sqrt{4-x}+\frac32.\] There are no other required dilations.

Now let's consider reflections. Looking at the rule of \(g\) we see that we need to implement a negative sign in front of the square root. So, we apply a reflection in the \(x\)-axis, giving \[f_2(x)=-f_1(x)=-\left(\sqrt{4-x}+\frac32\right)=-\sqrt{4-x}-\frac32.\]We also have an unwanted negative sign in front of the \(x\), so we should apply a reflection in the \(y\)-axis too, so that \[f_3(x)=f_2(-x)=-\sqrt{4-(-x)}-\frac32=-\sqrt{x+4}-\frac32.\]Now, let's consider translations. The inside of the square root in the rule of \(g\) is a lone \(x\). So, we apply a translation of \(4\) units in the positive \(x\)-direction: \[f_4(x)=f_3(x-4)=-\sqrt{(x-4)+4}-\frac32=-\sqrt{x}-\frac32.\]

Finally, we need to shift the graph up \(6-(-3/2)=15/2\) units: \[f_5(x)=f_4(x)+\frac{15}{2}=\left(-\sqrt{x}-\frac32\right)+\frac{15}{2}=-\sqrt{x}+6.\]All together, we end up with the same transformations. Ie: \[g(x)=\frac{-1}{2} f\big(\!-\!(x-4)\big)+\frac{15}{2}.\]

Either method works. Although Method 2 looks longer, it can be a lot quicker when given simpler problems involving only a few transformations.