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directions: identify the zeros, their multiplicity, an\ 1. $f(x) = x^2(…

Question

directions: identify the zeros, their multiplicity, an\

  1. $f(x) = x^2(x - 1)^4(x + 5)$\

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zeromultiplicityeffect\
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Explanation:

Step1: Find zeros from factors

To find the zeros of the polynomial \( f(x) = x^2(x - 1)^4(x + 5) \), we set each factor equal to zero.

  • For the factor \( x^2 \), set \( x = 0 \).
  • For the factor \( (x - 1)^4 \), set \( x - 1 = 0 \), so \( x = 1 \).
  • For the factor \( (x + 5) \), set \( x + 5 = 0 \), so \( x = -5 \).

Step2: Determine multiplicity

The multiplicity of a zero is the exponent of its corresponding factor.

  • For \( x = 0 \), the factor is \( x^2 \), so multiplicity is \( 2 \) (even multiplicity).
  • For \( x = 1 \), the factor is \( (x - 1)^4 \), so multiplicity is \( 4 \) (even multiplicity).
  • For \( x = -5 \), the factor is \( (x + 5)^1 \) (since there's no written exponent, it's 1), so multiplicity is \( 1 \) (odd multiplicity).

Step3: Determine the effect on the graph

  • For a zero with even multiplicity: The graph touches the x - axis and turns around at that zero (since the sign of the function doesn't change as it passes through the zero).
  • For a zero with odd multiplicity: The graph crosses the x - axis at that zero (since the sign of the function changes as it passes through the zero).
  • For \( x = 0 \) (multiplicity 2, even): Graph touches x - axis and turns around.
  • For \( x = 1 \) (multiplicity 4, even): Graph touches x - axis and turns around.
  • For \( x = -5 \) (multiplicity 1, odd): Graph crosses x - axis.

Answer:

ZeroMultiplicityEffect
\( 1 \)\( 4 \)Graph touches x - axis and turns around
\( -5 \)\( 1 \)Graph crosses x - axis