Question 175772
Let p(x)=x(x-3)^2(x+1)
A) Sketch this graph of p(x). Label all intercepts. <-- 
{{{graph(400,300,-3,4,-10,10,(x-3)^2(x^2+x))}}}
x-intercepts: x = 0, x= 3 with multiplicity 2, x= -1
y-intercept: P(0) = 0(0-3)^2(0+1) = 0

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b)Find another polynomial funtcion, q(x), that has the same zeroes as p(x) and goes throught the point (-1,16). <--- once i get the first graph I'm not sure How i would MAKE it go through (-1,16), How would i go upon doing that? 
Consider q(x) =  x(x-3)^2(x+1)+16
This has the same zeroes as p(x)
and q(-1) = 16
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c) Explain how to determine the end behavior of a polynomial function. <--- what does it mean by determine if we already solved it above?
"end behavior" refers to the y-values as x gets increasingly large
Since the highest power term determines that y value look 
at x(x-3)^2(x+1) = x^4+....
As x gets larger x^4 gets larger so y gets larger and the 
end behavior is "y gets increasingly larger".
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Cheers,
Stan H.