fx(x)=(x-2)(x-1)^2 find the zeros and multiplicity of each

Question

anonymous · Answer

Well the zeros are already given to you by the (x-2)(x-1)^2. Anything in the parentheses are the zeros if you set it equal to zero. For example, if you had (x-3) in your equation, set it equal to 0. x-3=0, you'd add three and you'd get x=3. So basically, the graph hits at 3,0. As for the multiplicity, that is the amount of times you have a certain zero. So if it was (x-3)^3(x+1), you'd have it hit at 3,0 3 times (because it's ^3) and it would hit at -1,0 once because it doesn't have any number in the exponent.

anonymous · Answer

Given that information can you figure out the zeros and multiplicity of your problem?

anonymous · Answer

Welcome to OpenStudy @barbaram !

barbaram · Answer

please explain the answer in mathematical form , for fx(x)=9x-2)(x-1)^2 find the zeros and multiplicitly ,find the number of turning points, where the graph cross the axis