f(x) = x^3 - 3x^2 - 9x + 6 x values: -3,-2,-1,0,1,2,3,4

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We have the polynomial function as y=x^3-3x^2-9x+6

To graph the function,we have to find the point of maxima and minima.

Differentiate the function with respect to x and equate it to zero,

d[f(x)]/dx=d(x^3-3x^2-9x+6)/dx=3x^2-6x-9 =0

x^2-2x-3=0

Solve for x

x^2-3x+x-3=0

x(x-3)+1(x-3)=0

(x+1)(x-3)=0

x=-1 and x=3

To determine whether maxima or minima occurs differentiate again and substitute the critical value of x

if d^2y/dx^2 >0 then minima occurs

and d^2/dx^2 <0 ,maxima occurs

Now,d^2y/dx^2 =d(f'(x)/dx=d(x^2-2x-3)/dx=2x-2

At x=-1 d^2y/dx^2 <0 therefore maxima occurs and maximum value is f(-1)=11

at x=3 minima ocur as d^2y/dsx^2 >0 ,At x=3 f(3)=-21

Also the y intercept are (0,6)

Using the above information creating a table of values

x y

-3 -21

-2 4

-1 11(maxima)

0 6

1 -5

2 -16

3 -21(minima)

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