# Graph the following function using the techniques of shifting, compressing, stretching, and/or reflecting.Start with the basic function and graph all stages on a single coordinate plane.`f(x)=...

Graph the following function using the techniques of shifting, compressing, stretching, and/or reflecting.

Start with the basic function and graph all stages on a single coordinate plane.`f(x)= -sqrt(x-5)`

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### 1 Answer

So, let's start with our the function we're trying to find:

`f(x) = -sqrt(x-5)`

What we're going to do is start by graphing a very basic function that looks a lot like that. How about:

`f(x) = sqrt(x)`

The graph is shown below:

So, we're looking at a function that "starts" at the origin and goes off to infinity.

Now, let's change this function slightly to get closer to the function we want. Let's let f(x) now be:

`f(x) = sqrt(x-5)`

What will this do? Well, now, at `x=0, f(x) = sqrt(-5)` so, clearly the function will no longer start at zero. Where will it start then? Where f(x) = 0. So, the function "starts," you could say, at x=5.

We have pretty much shifted the origin for our function 5 to the right compared to the previous graph! Here's the new graph of

`f(x) = sqrt(x-5)`

Hopefully, you can compare the two graphs and see that this graph is simply the one before, just shifted to the right 5 units.

Now to find the last part. Let's finally change our function into the negative of the one before:

`f(x) = -sqrt(x-5)`

This now gives us the original function. Notice, though, that we're basically taking the negative of the graph above.

What this means is that for every y-value associated with the above graph, the correct value will be -y. In other words, this will reflect the graph across the x-axis, seen here:

So, as a recap:

1) (Red) We start with the basic function

2) (Purple) We then shift it to the right 5 (because of the (x-5) term)

3) (Blue) We then reflect over the x-axis (because of the - term in front of the square root).

Shown here on a single graph:

I hope that helps!

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