Graphing a cubic function examples
      Graphing a cubic function type 2/1
The graphs of the cubic functions
There are three types (shapes of the graphs) of cubic functions:
type 1 y = a3x3 + a2x2 + a1x + a0    or    y - y0 = a3(x - x0)3,    - a22 + 3a3a1 = 0 or a1 = 0.
therefore, its source function  y = a3x3,  and the tangent line through the point of inflection is horizontal.
type 2/1 y = a3x3 + a2x2 + a1x + a0     or      y - y0 = a3(x - x0)3 + a1(x - x0), where  a3a1> 0
whose slope of the tangent line through the point of inflection is positive and equals a1.
type 2/2 y = a3x3 + a2x2 + a1x + a0     or      y - y0 = a3(x - x0)3 + a1(x - x0), where  a3a1< 0
whose slope of the tangent line through the point of inflection is negative and is equal a1
Graphing translated cubic function type 2/1
type 2/1 y = a3x3 + a2x2 + a1x + a0     or    y - y0 = a3(x - x0)3 + a1(x - x0),    a3  a1 > 0,
  I (x0, y0).
  Example:      Find the coordinates of translations, the zero point, the point of inflection and draw graphs of the cubic function  y = - x3 + 3x2 -  5x + 6 and its source function.
Solution:  1)  Calculate the coordinates of translations
        y0 f(x0)   =>   y0 = f(1) = -13 + 3 12 - 5 1 + 6  = 3,                   y0 = 3
Therefore, the point of inflection I (1, 3).
2)  To get the source cubic function, plug the coordinates of translations into the general form of the cubic,
y + y0 =  a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
thus,      y + 3 = -1 (x + 1)3 + 3 (x + 1)2 - 5 (x + 1) + 6    =>    y = - x3 - 2the source function.
Let prove that the source cubic is an odd function, which means that  f(x) = -f(-x),
since    f (x) = - x3 - 2then  -f (-x) = - [- (-x)3 - 2(-x)] = - [ x3 + 2x] = - x3 - 2x.
As,  a3 = - 1 and  a1 = - 2  then  a3a1 > 0  therefore, given function is of the type 2/1.
Since given function is symmetric to its point of inflection, and as the y-intercept a0 = 6, then the x-intercept or zero of the function must be at the point (2, 0) as shows the below figure.
To prove the root plug the values, a3 = - 1, a1 = - 2, x0 = 1 and y0 = 3 into  
so  x1 = 2,  therefore   - x3 + 3x2 -  5x + 6 = -1(x - 2)(x2 - x + 3).
Pre-calculus contents E
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