Third-degree, with zeros of −3 , − 2 , and 1 , and passes through the point ( 3 , 11 ) .
construct a polynomial function with the stated properties. reduce all fractions to lowest terms.

y  =   ( x + 3 ) *  ( x +  2 ) * ( x - 1 )* 11/60

Step-by-step explanation:

Lets   y  =  f(x)   in a cartesian coordinates

Having three zeros:

x = -3       ⇒   x  = -2      ⇒   and   x = 1  

That meas that if   x  takes the above mentioned  values " y " must be zero

therefore  y  must be of the form

f (x)  =  y  =  ( x + 3 ) *  ( x +  2 ) * ( x - 1 )     (1)

In that case for y to be zero one of the factors should be zero or

y  =  0        x  + 3 = 0    and  x = - 3  is a zero of the function y .

The same reasoning applies for the other two roots

Now we have to evaluate the other condition.

According to problem statement the function passes through the point ( 3, 11 ) , that means that when x =  3 ,  y have to be 11, therefore we plug in equation (1)  that value to see what happens

y  =   ( x + 3 ) *  ( x +  2 ) * ( x - 1 )

11  =  ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1)   =  6*5*2  = 60

Then we adjust the expression (1) to meet the condition of the function passing through point  ( 3 , 11) as:

y  =  ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1) * 11/60    (2)

and check to see if we did right

for y to be zero      x   can be    x = - 3    x  =  -2   and  x = 1 in all these cases y = 0.  And if  x  =  3   in equation (2)  y = 11. And that what we want to shown. Then the solution is:

y  =   ( x + 3 ) *  ( x +  2 ) * ( x - 1 )* 11/60

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64