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Question
f(x) = (x - 2)^3(x + 8)^2(x + 12)
at x = 2, the graph crosses the x-axis.
at x = -8, the graph touches the x-axis.
at x = -12, the graph crosses the x-axis.
complete
g(x) = (x + 4)^2(x - 9)
at x = -4, the graph select the x-axis.
at x = 9, the graph select the x-axis.
⚡ Using what you learned: Zeros and Multiplicity
Step 1: Analyze the zero at \(x = -4\)
The factor corresponding to \(x = -4\) is \((x + 4)^2\).
The exponent (multiplicity) is \(2\), which is an even number. When a zero has an even multiplicity, the graph touches the \(x\)-axis and turns around.
- At \(x = -4\), the graph touches the \(x\)-axis.
Step 2: Analyze the zero at \(x = 9\)
The factor corresponding to \(x = 9\) is \((x - 9)\), which can be written as \((x - 9)^1\).
The exponent (multiplicity) is \(1\), which is an odd number. When a zero has an odd multiplicity, the graph crosses the \(x\)-axis.
- At \(x = 9\), the graph crosses the \(x\)-axis.
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- At \(x = -4\), the graph touches the \(x\)-axis.
- At \(x = 9\), the graph crosses the \(x\)-axis.