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I had a question about $y=\sqrt{-x+4}$. I missed it, because I first reflected about the y axis then moved the graph left 4. That was wrong, because apparently I need to factor out a -1 to get $y=\sqrt{-1(x-4)}$, and the graph should actually shift to the right, instead of left.

Then I had $y=-2(3)^{5-x}$. Following the logic of the previous problem, I would factor out a -1 to get $y=-2(3)^{-1(-5+x)}$, and shift the graph to the right. However, this is not correct, and this graph is shifted to the left.

I don't get the difference between these two problems, when they both have a negative x, and a positive integer. Why does one shift to the left, and one shifts to the right?

Linda
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  • What was the question? – Sassatelli Giulio Aug 21 '22 at 18:41
  • In both the cases, the graph indeed shifts right. (Not left in the latter case). – VoidGawd Aug 21 '22 at 18:48
  • InanimateBeing, I now see that you are correct, and I'm not sure why I thought the second one was shifted left. I think I just needed a break from this topic, and perhaps more practice and thought about the order of operations. – Linda Aug 21 '22 at 19:02
  • In your second example $y=-2(3)^{5-x}$ think of it as replacing $y=2(3)^{5-x}$ with $-y=2(3)^{5-x}$. When $y$ is replaced with $-y$ the graph is reflected about the $x$-axis. All examples of graphing $y=-f(x)$ should be conceptualized as replacing $y$ with $-y$. Then it is clear that the logic is the same as when replacing $x$ with $-x$. Otherwise one is left with a semblance of inconsistency in the operations. – John Wayland Bales Aug 22 '22 at 00:09
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1 Answers1

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Let's do the graph transformation for both (starting from step $0$): $$y=\sqrt x$$
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$$y=\sqrt {-x}$$
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$$y=\sqrt {-(x-4)}$$
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$$y=K^x$$
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$$y=-K^x$$
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$$y=-K^{-x}$$
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$$y=-K^{-(x-5)}$$
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In case you ever have confusion, you may refer this link on Graph Transformation.

VoidGawd
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