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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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