Polynomial and/or Polynomial Functions and Equations
      Quadratic function and equation
         The graph of quadratic function
         Translated form of quadratic function
         The sum and product of the roots
         Vertex (the turning point, maximum or minimum) - coordinates of translations
         Transformations of the graph of the quadratic function
      Quadratic equation word problems
Quadratic function and equation
The polynomial function of the second degree,  f (x) = a2x2 + a1x + a0 is called a quadratic function.
   y = f (x = a2x2 + a1x + a0   or   y - y0 = a2(x - x0)2,  where
coordinates of translations of the quadratic function. 
By setting  x0 = y0 = 0,  we get y = a2x2,  the source quadratic function.  The turning point  V (x0, y0).
The real zeros of the quadratic function:  
   y = f (x= a2x2 + a1x + a0  = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2]
The graph of a quadratic function is curve called a parabola. The parabola is symmetric with respect to a vertical line called the axis of symmetry.
As the axis of symmetry passes through the vertex of the parabola its equation is x = x0.
Transformations of the graph of the quadratic function
How changes in the expression of the quadratic function affect its graph is shown in the figures below.
  The graph of quadratic polynomial will intersect the x-axis in two distinct points if its leading coefficient a2 and the vertical translation y0 have different signs, i.e., if   a2 y0 < 0.
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 = + sqrt(4)
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 explored under the title, 'Transformations of the polynomial function applied to the quadratic and cubic functions' 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.
Quadratic equation word problems
Example:  A train made up for delay of 12 minutes after 60 km of way by running 10 km/h faster then regular speed.  What is the regular speed of the train? 
Solution:
 
 
 
 
Example:  In a theater each row has the same number of seats. Number of rows equals number of seats. By doubling number of rows and decreasing number of seats 10 per row, total number of seats in the theater increases by 300. How many rows are in the theater?
Solution:  Taking x as the number of rows (or seats per row), then
College algebra contents C
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