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Quadratic Functions by In3PZQE6

VIEWS: 4 PAGES: 19

									      Sketching quadratic functions
To sketch a quadratic function we need to identify where possible:


The shape:   ax 2  bx  c

If a  0 then                   If a  0 then

The y intercept (0, c)

The roots by solving ax2 + bx + c = 0

The axis of symmetry (mid way between the roots)

The coordinates of the turning point.
1. Sketch the graph of y  5  4 x  x 2

The shape                           2– 2
                                    6 6
                                    4 4
                                    10 10
                                    8 8
                                    2– 2
                                    4 4
                                    10 10
                                    8 8
                                    6 6

 The coefficient of x2 is -1 so the shape is
The Y intercept

 (0 , 5)                                                                      y
                                                          (-2 , 9)      10
                                                                         8
 The roots
                                                                         6

 5  4x  x  0
              2
                                                                         4

(5  x)(1  x)  0                                                       2

                                               – 10 – 8   – 6   – 4   – 2         2   4   6   8   10   x
                                                                        – 2
(-5 , 0) (1 , 0)
                                                                        – 4

The axis of symmetry                                                    – 6
                                                                        – 8

Mid way between -5 and 1 is -2                                         – 10

 x = -2
The coordinates of the turning point
When x  2, y  9
 (-2 , 9)
               Completing the square
 The coordinates of the turning point of a quadratic can also be found by completing
 the square.
 This is particularly useful for parabolas that do not cut the x – axis.


REMEMBER

y  ax 2  bx  c can be written in the form y  a( x  p ) 2 +q.
Axis of symmetry x   p
Coordinates of turning point (  p, q).
1. Find the equation of the axis of symmetry and the coordinates of the
turning point of y  2 x 2  8 x  9.
 y  2 x2  8x  9
    2( x 2  4 x)  9
    2( x  2) 2  9  8
    2( x  2)2  1

Axis of symmetry is x = 2
Coordinates of the minimum turning point is (2 , 1)
2. Find the equation of the axis of symmetry and the coordinates of the
turning point of y  7  6 x  x 2 .
 y   x2  6x  7
    ( x 2  6 x )  7
    ( x  3) 2  7  9
    ( x  3)2  16

Axis of symmetry is x = 3
Coordinates of the maximum turning point is (3 , 16)
       Solving quadratic equations
  Quadratic equations may be solved by:
      The Graph
      Factorising
      Completing the square
      Using the quadratic formula


1. Solve 6 x 2  x  15  0
     (2 x  3)(3 x  5)  0

 2x  3  0         3x  5  0
          3                  5
     x                  x
          2                  3
2. Solve x 2  4 x  1  0
This does not factorise.
                              b  b 2  4ac
ax 2  bx  c  0          x
                                   2a
a  1, b  4, c  1

   4  16  4
x
         2
   4  20
 
       2
   42 5
 
      2
   2(2  5)
 
       2
  2  5 or 2  5
                 Quadratic inequations
 A quadratic inequation can be solved by using a sketch of the quadratic function.

1. For what values of x is 12  5 x  2 x 2  0?
First do a quick sketch of the graph of the function.

                                    12  5 x  2 x 2  (4  x)(3  2 x)
                                    Roots are -4 and 1.5
                                    Shape is a

  -4       1.5                      The function is positive when it is above the x axis.

                                              3
                                     4  x 
                                              2
2. For what values of x is 12  5 x  2 x 2  0?
First do a quick sketch of the graph of the function.

                                    12  5 x  2 x 2  (4  x)(3  2 x)
                                    Roots are -4 and 1.5
                                    Shape is a

  -4       1.5                      The function is negative when it is below the x axis.

                                                  3
                                   4  x and x 
                                                  2
              The quadratic formula
 ax 2  bx  c  0

   b  b 2  4ac
x
        2a
1. Solve 2 x 2  5 x  1  0 compare with ax  bx  c  0
                                            2


   a  2, b  5, c  1
     5  25  8
  x
          4
     5  17       5  17
             and
        4            4
     2.28 and 0.22
 2. Solve x 2  6 x  9  0   compare with ax 2  bx  c  0
   a  1, b  6, c  9
      6  36  36
   x
          2
      60
    
        2
      3


From the above example when the number under the square root sign is zero there
is only 1 solution.
  3. Solve x 2  2 x  9  0   compare with ax 2  bx  c  0
     a  1, b  2, c  9
      2  4  36
   x
            2
      2  32
    
          2
Since 32 is not a real number there are no real roots.

From the above example we require the number under the square root sign to be
positive in order for 2 real roots to exist.
   This leads to the following observation.

   If b 2  4ac  0 the roots are real and unequal.
   If b 2  4ac  0 the roots are real and equal.
   If b 2  4ac  0 the roots are non-real.

   For the quadratic equation ax 2  bx  c  0,
   b 2  4ac is called the DISCRIMINANT.

4. Find the nature of the roots of 4 x 2  12 x  9  0
  a  4, b  12, c  9
  b 2  4ac  144  144
            0
Since the discriminant is zero, the roots are real and equal.
                Using the discriminant
 We can use the discriminant to find unknown coefficients in a quadratic equation.

1. Find p given that 2 x 2  4 x  p  0 has real roots.
  a  2, b  4, c  p
  b 2  4ac  16  8 p  0
                    8 p  16
                      p2

The equation has real roots when p  2.
3. Show that the roots of (k  2) x 2  (3k  2) x  2k  0 are always real.
  a  (k  2), b  (3k  2), c  2k
                  2  3k

  b2  4ac  (4  12k  9k 2 )  8k (k  2)
             4  12k  9k 2  8k 2  16k
             k 2  4k  4
             (k  2) 2
              (k  2) 2  0 for all values of k .

  Since the discriminant is always greater than or equal to zero, the roots of the
  equation are always real.
            Conditions for tangency
To determine whether a straight line cuts, touches or does not meet a curve the
equation of the line is substituted into the equation of the curve.

When a quadratic equation results, the discriminant can be used to find the number
of points of intersection.



 If b 2  4ac  0 there are two distinct points of intersection.

 If b 2  4ac  0 there is only one point of intersection.
                    the line is a tangent to the curve.
 If b 2  4ac  0 the line does not intersect the curve.
1. Prove that the line y  2 x  1 is a tangent to the curve y  x 2 .
   Find the point of intersection.

           x2  2x 1
  x2  2 x  1  0
  a  1, b  2, c  1
  b 2  4ac  4  4  0
  Since the discriminant is zero, the line is a tangent to the curve.

  x2  2 x  1  0
      ( x  1) 2  0
              x 1     y  x2  1


 Hence the point of intersection is (1 , 1).
2. Find the equation of the tangent to y  x 2  1 that has gradient 2.

 A straight line with gradient 2 is of the form y  2 x  k
   x2  1  2x  k
  x 2  2 x  (1  k )  0
   a  1, b  2, c  1  k

   For tangency b 2  4ac  0
             4  4(1  k )  0
                  4  4  4k  0
                          4k  0
                           k 0

 Hence the equation of the tangent is y = 2x.
3. Find the equations of the tangents from (0,-2) to the curve y  8 x 2 .

 A straight line passing through (0,-2) is of the form y  mx  2
            8 x 2  mx  2                                y


   8 x 2  mx  2  0

   a  8, b   m, c  2
  For tangency b 2  4ac  0
                                                                         x

                 m 2  64  0
                      m 2  64
                        m  8


Hence the equation of the two tangents are y = 8x – 2 and y = -8x - 2.

								
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