Polynomial and/or Polynomial Functions and Equations Graphing polynomial functions
Cubic functions
Graphs of cubic functions
Cubic functions
Using the same method we can analyze the third degree polynomial or cubic functions.
y = a3x3 + a2x2 + a1x + a0     or      y - y0 = a3(x - x0)3 + a1(x - x0), By setting  x0 = 0  and  y0 = 0 we get the source cubic function  y = a3x3 + a1x  where  a1= tanat
Coordinates of the point of inflection coincide with the coordinates of translations, i.e.,  I (x0, y0).
The source cubic functions are odd functions.
Graphs of odd functions are symmetric about the origin that is, such functions change the sign but not absolute value when the sign of the independent variable is changed, so that  f (x) = - f (-x).
That is, change of the sign of the independent variable of a function reflects the graph of the function about the y-axis, while change of the sign of a function reflects the graph of the function about the x-axis.
The graphs of the translated cubic functions are symmetric about its point of inflection.
As shows the figure below there are three types of cubic functions:
 type 1 y = a3x3 + a2x2 + a1x + a0     or      y - y0 = a3(x - x0)3   where 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 equal 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
 The graph of its source function has three zeros and two turning points at Graphs of cubic functions 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 = -(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),
As  f (x) = - x3 - 2then  - f (-x) = - [- (-x)3 - 2(-x)] = - [ x3 + 2x] = - x3 - 2x.
As a3a1 > 0 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).
Therefore,   - x3 + 3x2 -  5x + 6 = -1(x - 2)(x2 - x + 3)    Intermediate algebra contents 