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 notationFirst, 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 intuitionFor 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.