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ALGEBRA
 Graphs of polynomial functions
 :: Quadratic function, the second degree polynomial   f (x) = a2x2 + a1x + a0
  By plugging the coordinates of translations into the source quadratic function  f (x) = a2x2  we get
f (x) = y = a2x2 + a1x + a0     or    y - y0a2(x - x0)2
By setting  x0 = 0  and  y0 = 0  obtained is the source quadratic  y a2x2 .
If  a2 y0 < 0  then the roots are   The turning point  V (x0 , y0 ).
 :: Quadratic equation, quadratic formula
The zeros of a polynomial function are the values of x for which the function equals zero.
Therefore, the solutions of the equation  f (x) = 0, that are called roots of the polynomial, are the zeros of the polynomial function or the x-intercepts of its graph in a coordinate plane. Let solve the equation,
a2x2 + a1x + a0 = 0     or    ax2 + bx + c = 0
 
 :: Cubic function, the third degree polynomial   f (x) = a3x3 + a2x2 + a1x + a0
By plugging the coordinates of translations into the source cubic function  f (x) = a3x3 + a1x  we get
f (x) = a3x3 + a2x2 + a1x + a0     or    y - y0a3(x - x0)3 + a1(x - x0)  where,
by setting  x0 = 0  and  y0 = 0 we get the source cubic function  y = a3x3 + a1x  where a1= tan at .
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.

There are three types (shapes) of cubic functions whose graphs of the source functions are shown in the figure below:

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
and 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.
The graph of the source function type 2/2 has three zeros or roots at
and two turning points at
The graphs of the source cubic functions - the classification criteria diagram
Translated cubic functions
type 1
 y = a3x3 + a2x2 + a1x + a0   or   y - y0 = a3(x - x0)3,   - a22 + 3a3a1 = 0 and where 
The root
 The point of inflection  I(x0, y0).
type 2/1
 y = a3x3 + a2x2 + a1x + a0   or    y - y0 = a3(x - x0)3 + a1(x - x0)  where,  a3a1> 0
The point of inflection  I(x0, y0).
The root
type 2/2
 y = a3x3 + a2x2 + a1x + a0  or   y - y0 = a3(x - x0)3 + a1(x - x0)  where  a3a1< 0
If  | y0 | > | yT |  
If  | y0 | < | yT |  
The turning points   The point of inflection  I(x0, y0).
 :: Drawing a cubic function example
Example:  Given is cubic function y = (- 1/3)x3 - 4x2 - 12x - 25/3, find its source or original function and calculate the coordinates of translations, the zero points, the turning points and the point of inflection. Draw graphs of the source and the given cubic.
Solution:  1)  Let calculate the coordinates of translations,
y0 f (x0),     y0 = f (- 4) = (-1/3) (- 4)3 - 4 (- 4)2 - 12 (- 4) - 25/3  = - 3,    y0 = - 3.
Therefore, the point of inflection I (- 4, - 3).
2)  To get the source cubic function we plug the coordinates of translations (with changed signs) into the general form of cubic,
y + y0 =  a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0   thus,
y - 3 = (-1/3) (x - 4)3 - 4(x - 4)2 - 12(x - 4) - 25/3,    y = (-1/3)x3 + 4the source function.
As,  a3 < 0  and  a3a1 < 0  given cubic is the type 2/2 whose graph is reflected around the x-axis.
then, the function has three real roots,
By plugging the coefficients of the source function and the coordinates of translations x0 and y0 into the above formula, we get the turning points,
 
 
 
 
 
 
 
 
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