Graph the function.

`y = 1/(x+3) + 3`

Identify the domain and range.

And, compare the graph with the graph of `y=1/x` .

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(I) Graph.

The graph of `y=1/(x+3) + 3` is:

Base on the graph, the shape of the two curves are the same. Their difference is that the curves of y=1/(x+3)+3 are moved three units to the left and three units up.

(II) Domain

Notice that the graph of the function has two separate curves.

For the curve at the left, its left end goes continuously to the left without bound. And its right end approaches the x at -3, but does not cross x=-3.

For the curve at the right, its left end approaches the the x at -3, but never crosses x=-3 too. And, its right end goes continuously to the right without bound.

**Hence, the domain of the given function is `(-oo, -3)uu(-3,+oo)` .**

(III) Range

The range refers to the y values of the function. So, consider the vertical direction of the curve.

For the left curve, its left end approaches y at 3, but does not pass through y=3. And is right end goes continuously down without bound.

While the right curve, its left end goes continuously up without bound. And its right end approaches y at 3 but never crosses.

**Thus, the range is `(-oo, 3)uu(3,+oo)` .**

(IV) Compare the graph given function with y=1/x.

The graph of y=1/x.

Base on the graph, the shape of the two curves are the same. Their difference is that the curves of `y=1/(x+3)+3` are moved three units to the left and three units up.

*Please take note of this correction*.

The paragraph that I have indicated here should not be on first part of the problem. It should only be in the last part since it compares the graph of the two functions.

Base on the graphs, the shape of the two curves `(y=1/x` and

`y=1/(x+3)+3)` are the same. Their difference is that the

curves of `y=1/(x+3)+3` ` `are moved three units to the left and three units up.

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