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Chapter 2: Problem 27

Find the zeros for each polynomial function and give the multiplicity for each zero. State whether the graph crosses the \(x\) -axis, or touches the \(x\) -axis and turns around, at each zero. $$f(x)=4(x-3)(x+6)^{3}$$

Short Answer

Expert verified
The zeros of the polynomial function \(f(x) = 4(x - 3)(x + 6)^{3}\) are \(x = 3\) and \(x = -6\). The zero \(x = 3\) has a multiplicity of 1 and the graph crosses the \(x\)-axis at this point. The zero \(x = -6\) has a multiplicity of 3 and the graph also crosses the \(x\)-axis at this point.

Step by step solution

01

Find the Zeros

Set the function equal to zero and solve for \(x\). Therefore, we get \(4(x - 3)(x + 6)^{3} = 0\). Setting each factor to zero gives two solutions, \(x - 3 = 0\) yields \(x = 3\) and \(x + 6 = 0\) gives \(x = -6\).
02

Find the Multiplicity

The exponent on each term indicates the multiplicity of the zero. From the equation \(4(x - 3)(x + 6)^{3} = 0\), the zero \(x = 3\) has multiplicity 1 due to the power on (x - 3) and \(x = -6\) has multiplicity 3 due to the power on (x + 6)^{3}.
03

Determine the Behavior of the Graph

If the multiplicity of the zero is odd, then the graph crosses the \(x\)-axis at that point. If it is even, the graph touches the \(x\)-axis at that point and turns around. In this case, \(x = 3\) has multiplicity 1 which is odd, hence the graph crosses the \(x\)-axis at \(x = 3\). \(x = -6\) has a multiplicity 3 which is also odd, so the graph crosses the \(x\)-axis at \(x = -6\) as well.

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Find the horizontal asymptote, if there is one, of the graph of rational function. $$g(x)=\frac{15 x^{2}}{3 x^{2}+1}$$

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