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g ( x ) = ( x - 1 ) 2 + 2 y i n t = ( 0 - 1 ) 2 + 2 = ( - 1 ) 2 + 2 = 1 + 2 = 3

The x -intercepts are calculated as follows:

y = a ( x + p ) 2 + q 0 = a ( x i n t + p ) 2 + q a ( x i n t + p ) 2 = - q x i n t + p = ± - q a x i n t = ± - q a - p

However, [link] is only valid if - q a > 0 which means that either q < 0 or a < 0 but not both. This is consistent with what we expect, since if q > 0 and a > 0 then - q a is negative and in this case the graph lies above the x -axis and therefore does not intersect the x -axis. If however, q > 0 and a < 0 , then - q a is positive and the graph is hat shaped with turning point above the x -axis and should have two x -intercepts. Similarly, if q < 0 and a > 0 then - q a is also positive, and the graph should intersect with the x -axis twice.

For example, the x -intercepts of g ( x ) = ( x - 1 ) 2 + 2 are given by setting y = 0 to get:

g ( x ) = ( x - 1 ) 2 + 2 0 = ( x i n t - 1 ) 2 + 2 - 2 = ( x i n t - 1 ) 2

which has no real solutions. Therefore, the graph of g ( x ) = ( x - 1 ) 2 + 2 does not have any x -intercepts.

Intercepts

  1. Find the x- and y-intercepts of the function f ( x ) = ( x - 4 ) 2 - 1 .
  2. Find the intercepts with both axes of the graph of f ( x ) = x 2 - 6 x + 8 .
  3. Given: f ( x ) = - x 2 + 4 x - 3 . Calculate the x- and y-intercepts of the graph of f .

Turning points

The turning point of the function of the form f ( x ) = a ( x + p ) 2 + q is given by examining the range of the function. We know that if a > 0 then the range of f ( x ) = a ( x + p ) 2 + q is { f ( x ) : f ( x ) [ q , ) } and if a < 0 then the range of f ( x ) = a ( x + p ) 2 + q is { f ( x ) : f ( x ) ( - , q ] } .

So, if a > 0 , then the lowest value that f ( x ) can take on is q . Solving for the value of x at which f ( x ) = q gives:

q = a ( x + p ) 2 + q 0 = a ( x + p ) 2 0 = ( x + p ) 2 0 = x + p x = - p

x = - p at f ( x ) = q . The co-ordinates of the (minimal) turning point is therefore ( - p , q ) .

Similarly, if a < 0 , then the highest value that f ( x ) can take on is q and the co-ordinates of the (maximal) turning point is ( - p , q ) .

Turning points

  1. Determine the turning point of the graph of f ( x ) = x 2 - 6 x + 8 .
  2. Given: f ( x ) = - x 2 + 4 x - 3 . Calculate the co-ordinates of the turning point of f .
  3. Find the turning point of the following function by completing the square: y = 1 2 ( x + 2 ) 2 - 1 .

Axes of symmetry

There is only one axis of symmetry for the function of the form f ( x ) = a ( x + p ) 2 + q . This axis passes through the turning point and is parallel to the y -axis. Since the x -coordinate of the turning point is x = - p , this is the axis of symmetry.

Axes of symmetry

  1. Find the equation of the axis of symmetry of the graph y = 2 x 2 - 5 x - 18 .
  2. Write down the equation of the axis of symmetry of the graph of y = 3 ( x - 2 ) 2 + 1 .
  3. Write down an example of an equation of a parabola where the y-axis is the axis of symmetry.

Sketching graphs of the form f ( x ) = a ( x + p ) 2 + q

In order to sketch graphs of the form f ( x ) = a ( x + p ) 2 + q , we need to determine five characteristics:

  1. sign of a
  2. domain and range
  3. turning point
  4. y -intercept
  5. x -intercept

For example, sketch the graph of g ( x ) = - 1 2 ( x + 1 ) 2 - 3 . Mark the intercepts, turning point and axis of symmetry.

Firstly, we determine that a < 0 . This means that the graph will have a maximal turning point.

The domain of the graph is { x : x R } because f ( x ) is defined for all x R . The range of the graph is determined as follows:

( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 - 3 - 3 f ( x ) - 3

Therefore the range of the graph is { f ( x ) : f ( x ) ( - , - 3 ] } .

Using the fact that the maximum value that f ( x ) achieves is -3, then the y -coordinate of the turning point is -3. The x -coordinate is determined as follows:

- 1 2 ( x + 1 ) 2 - 3 = - 3 - 1 2 ( x + 1 ) 2 - 3 + 3 = 0 - 1 2 ( x + 1 ) 2 = 0 Divide both sides by - 1 2 : ( x + 1 ) 2 = 0 Take square root of both sides: x + 1 = 0 x = - 1

The coordinates of the turning point are: ( - 1 , - 3 ) .

The y -intercept is obtained by setting x = 0 . This gives:

y i n t = - 1 2 ( 0 + 1 ) 2 - 3 = - 1 2 ( 1 ) - 3 = - 1 2 - 3 = - 1 2 - 3 = - 7 2

The x -intercept is obtained by setting y = 0 . This gives:

0 = - 1 2 ( x i n t + 1 ) 2 - 3 3 = - 1 2 ( x i n t + 1 ) 2 - 3 · 2 = ( x i n t + 1 ) 2 - 6 = ( x i n t + 1 ) 2

which has no real solutions. Therefore, there are no x -intercepts.

We also know that the axis of symmetry is parallel to the y -axis and passes through the turning point.

Graph of the function f ( x ) = - 1 2 ( x + 1 ) 2 - 3

Khan academy video on graphing quadratics

Sketching the parabola

  1. Draw the graph of y = 3 ( x - 2 ) 2 + 1 showing all the intercepts with the axes as well as the coordinates of the turning point.
  2. Draw a neat sketch graph of the function defined by y = a x 2 + b x + c if a > 0 ; b < 0 ; b 2 = 4 a c .

Writing an equation of a shifted parabola

Given a parabola with equation y = x 2 - 2 x - 3 . The graph of the parabola is shifted one unit to the right. Or else the y-axis shifts one unit to the left i.e. x becomes x - 1 . Therefore the new equation will become:

y = ( x - 1 ) 2 - 2 ( x - 1 ) - 3 = x 2 - 2 x + 1 - 2 x + 2 - 3 = x 2 - 4 x

If the given parabola is shifted 3 units down i.e. y becomes y + 3 . The new equation will be:

(Notice the x-axis then moves 3 units upwards)

y + 3 = x 2 - 2 x - 3 y = x 2 - 2 x - 6

End of chapter exercises

  1. Show that if a < 0 , then the range of f ( x ) = a ( x + p ) 2 + q is { f ( x ) : f ( x ) ( - , q ] } .
  2. If (2,7) is the turning point of f ( x ) = - 2 x 2 - 4 a x + k , find the values of the constants a and k .
  3. The graph in the figure is represented by the equation f ( x ) = a x 2 + b x . The coordinates of the turning point are (3,9). Show that a = - 1 and b = 6 .
  4. Given: y = x 2 - 2 x + 3 . Give the equation of the new graph originating if:
    1. The graph of f is moved three units to the left.
    2. The x -axis is moved down three units.
  5. A parabola with turning point (-1,-4) is shifted vertically by 4 units upwards. What are the coordinates of the turning point of the shifted parabola?

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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