For the polynomial, list each real zero and its multiplicity. Determine whether the graph crosses or touches the x-axis at each x-intercept.
f(x) = 5(x-5)(x-4)^4
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so what would be the zeros there?
keep in mind that \(\bf f(x) = 5(x-5)(x-4)^4\\ \implies f(x) = 5(x-5)(x-4)(x-4)(x-4)(x-4)\)
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yeap x = 5 and x = 4
but notice that you have (x-4) 4 times, that means "multiplicity"
if the multiplicity is ODD, that is an ODD EXPONENT, the graph crosses the x-axis at that point
if the multiplicity is EVEN, that is an EVEN EXPONENT, the graph touches the x-axis only
The zeros will be those x-values for which f(x) = 0. These would be 5 and 4, because x=5 sends (x-5) to zero, and therefore f(5) = 0, and similarly for x=4. The "multiplicity" of a zero is the exponent on it: (x-5)^1 has multiplicity 1, (x-4)^4 has multiplicity 4. If a zero has ODD multiplicity, it will cross the x-axis. If a zero has EVEN multiplicity, it will touch the x-axis, but will not cross it.
I'll leave it to you to think about WHY this is the case.
thanks a million, i think i understand it from her