Find the x intercepts and graph:

y = (x + 3)^2 (2x + 1) (2 - x)

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The given function is `y = (x + 3)^2 (2x + 1) (2 - x)`

For x intercepts, put y=0. Thus,

`(x + 3)^2 (2x + 1) (2 - x)=0`

For `(x + 3)^2=0` `x=-3` (double root)

For `(2x + 1)=0` `x=-1/2` (simple root)

For `(2 - x)=0` ` x=2` (simple root)

The graph of the function is :

Since, there is a simple root at -1/2 and 2 the graph crosses the x axis at` x=-1/2` and `x=2` .

The graph has a double root at `x=-3` , so it just touches the x axis at -3 but doesn't cross the axis. The curve has x axis as tangent at this point.

**Therefore, the x intercepts are -1/2 and 2, where the curve cuts the x axis. The curve touches x axis at x=-3.**

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