I have obtained all the values to graph f(x)= x^3-2x^2+x-2.

Here they are:

Assume a cubic function as f(x) = 23 - 2x2 + x - 2. Let y=f(x). Substitute x=0 into y = 23 - 2x2 + x - 2. y = 03 -2(0)2+0-2 -2 Substitute y=0 into y = x3 - 2x- + x - 2. 0 = 203 - 220 + 20 - 2 (2 - 2) (202 + 1) = 0 a - 2 = 0 or x +1=0 = 2, x = i, x = -i Ignore complex solutions. So, the x-intercept is (2,0) and the y-intercept is (0,-2). To find the critical point, determine the derivative of f(x). fi(ac) = (203) - 2 (2202) + 2 (20) - (2) = 3x2 - 4x +1 -0 3x2 - 4x + 1 Equate the obtained derivative to zero, and solve for x. 3x2 - 4x + 1 = 0 -(-4)+ (-4)2-43.1 = 2.3 4+V16-12 2.3 = 4+2 2.3 = 4+2 4-2 6 ' 6 = 6 2 6 ' 6 = iticHence the critical points are x=l, and X=V3. From the critical points, the intervals are (00, i 1 ), ($1), Make a table to check the sign of the derivative f'(x). f'(x) Interval (-00, %) (1,00) Hence the function f(x) is increasing in the interval (00, F X 07 Mar) 2 3-0.724(0.7)+1 2 f'(-1) : 3(1)2 4(1)+ 1 8 0. 33 f'(2) : 3(2)2 - 4(2) + 1 interval (%,1). 5 Sign (1, 00). + (increasing) (decreasing) + (increasing) 1 To nd the inflection points, nd the second derivative. f\"(:c) d Equate the second derivative to O and solve for X. 6324 Substitute a: f( 33 2 3 ) 0 g 3 ) U (1, co) and f(x) is decreasing in the To find the inflection points, find the second derivative. fli (20) = (3ac2 - 4x + 1 ) = d (3ac2) - d (4x) + + (1) 6x - 4 Equate the second derivative to O and solve for x. 6x - 4 = 0 Co/ N Substitute x = 3 into f(xc) = 23 - 2x2 + x - 2. f ( 3) = (3)3-2(2)2+(2) -2 = 27 - 2 . 8 + 2 - 2 -2.2.27-28 27 52 27 Hence the inflection point of f(x) is 52 Check the sign of the second derivative in the intervals (-oo, 2 ) and (3, co). Interval X f" (x ) (-0o, ?) -1 f"(-1) = 6(-1) -4 = -10 (3,00) 1 fi(1) = 6(1) -4 + 2 Hence the function is concave upward on the interval (2, co ) and the function is concave downward on the interval (-co, ?) Use the above-obtained values to graph the function f(x)