Sure, I can help you graph the function [tex]$y = -\sqrt{x - 1} + 5[/tex]$. Here are the steps involved:
1. Shift the graph of $y = -\sqrt{x}$ to the right by one unit. The graph of [tex]$y = -\sqrt{x - 1}$[/tex] is the same as the graph of $y = -\sqrt{x}$ shifted one unit to the right. To do this, imagine taking the graph of $y = -\sqrt{x}$ and moving all the points one unit to the right.
2. Reflect the graph of [tex]$y = \sqrt{x - 1}$[/tex] across the y -axis.** The graph of [tex]$y = -\sqrt{x - 1}$[/tex] is the reflection of the graph of [tex]$y = \sqrt{x - 1}$[/tex] across the y -axis. To do this, imagine taking the graph of [tex]$y = \sqrt{x - 1}$[/tex] and flipping it across the y -axis.
3. Shift the graph of [tex]$y = -\sqrt{x - 1}$[/tex] up five units. The graph of [tex]$y = -\sqrt{x - 1} + 5$[/tex] is the graph of [tex]$y = -\sqrt{x - 1}$[/tex] shifted up five units. To do this, imagine taking the graph of [tex]$y = -\sqrt{x - 1}$[/tex] and moving all the points five units up.
Once you have completed these steps, you should have a graph that looks like the one in the image you sent me.
Here are some additional things to keep in mind:
* The domain of the function is all numbers greater than or equal to 1, since the radicand (the expression under the radical) cannot be negative.
* The range of the function is all numbers less than or equal to 5, since the function is always less than or equal to 5 (due to the reflection across the y -axis) and it approaches 5 as x approaches positive infinity.
* The graph of the function has a hole at the point [tex]$(1, 5)$[/tex], since the function is undefined at [tex]$x = 1$[/tex].
I hope this helps!
