Polynomial and/or Polynomial Functions and Equations
Graphing polynomial functions
Zero polynomial
Constant function
Linear function
Transition of the quadratic polynomial from the general to source form and vice versa
The zeros or the roots of the quadratic function
Vertex (the turning point, maximum or minimum) - coordinates of translations
Graphing polynomial functions
Polynomial functions are named in accordance to their degree.
Zero polynomial   f (x) = 0
The constant polynomial  f (x) = 0 is called the zero polynomial and is graphically represented by the x-axis.
Constant function  f (x) = a0
A polynomial of degree 0,  f (x) = a0, is called a constant function, its graph is a horizontal line with
y-intercept a0.
Linear function   ya1x + a0
ya1x + a0  or  ya1(x - x0)  or  y - y0a1x,
where  x0 - a0/a1  and/or   y0a0.
By setting  x0 = 0  or  y0 = 0  obtained is
 the source linear  y = a1x.
Quadratic function    y = a2x2 + a1x + a0
1)  Let calculate the coordinates of translations of quadratic function using the formulas,
 substitute n = 2 in
 then
2)  To get the source quadratic function we should plug the coordinates of translations (with changed signs)
into the general form of the quadratic, i.e.,
after expanding and reducing obtained is
y = a2x2   the source quadratic function.
3)  Inversely, by plugging the coordinates of translations into the source quadratic function
y - y0 = a2(x - x0)2,
and after expanding and reducing we obtain
y = a2x2 + a1x + a0   the quadratic function in the general form.
ƒ(x) = y = a2x2 + a1x + a0   or   y - y0a2(x - x0)2,
 where
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 ).
Example:  Find zeros and vertex of the quadratic function  y = - x2 + 2x + 3  and sketch its graph.
Solution:  A quadratic function can be rewritten into translatable form  y - y0 = a2(x - x0)2  by completing the square,
 y = - x2 + 2x + 3 Since a2 · y0 < 0 given quadratic function must have two different real zeros. y = - (x2 - 2x) + 3 To find zeros of a function, we set y equal to zero and solve for x. Thus, y = - [(x - 1)2 - 1] + 3 - 4 = - (x - 1)2 y - 4 = - (x - 1)2 (x - 1)2 = 4 y - y0 = a2(x - x0)2 x - 1 = ± Ö4 V(x0, y0)  =>  V(1, 4) x1,2 = 1 ± 2,   =>    x1 = - 1 and  x2 = 3.
We can deal with given quadratic using the property of the polynomial shown above. Thus,
1)  calculate the coordinates of translations of the quadratic  y = f (x= - x2 + 2x + 3
2)  To get the source quadratic function, plug the coordinates of translations (with changed signs)
into the general form of the quadratic, i.e.,
y + y0 = a2(x + x0)2 + a1(x + x0) + a0   =>    y + 4 = - (x + 1)2 + 2(x + 1) + 3
y = - x2   the source quadratic function
3)  Inversely, by plugging the coordinates of translations into the source quadratic function
y - y0 = a2(x - x0)2   =>     y - 4 = - (x - 1)2
obtained is given quadratic in general form     y = - x2 + 2x + 3.
Pre-calculus contents E