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Functions of the form y = a b ( x ) + q

Functions of the form y = a b ( x ) + q are known as exponential functions. The general shape of a graph of a function of this form is shown in [link] .

General shape and position of the graph of a function of the form f ( x ) = a b ( x ) + q .

Investigation : functions of the form y = a b ( x ) + q

  1. On the same set of axes, plot the following graphs:
    1. a ( x ) = - 2 · b ( x ) + 1
    2. b ( x ) = - 1 · b ( x ) + 1
    3. c ( x ) = 0 · b ( x ) + 1
    4. d ( x ) = 1 · b ( x ) + 1
    5. e ( x ) = 2 · b ( x ) + 1
    Use your results to deduce the effect of a .
  2. On the same set of axes, plot the following graphs:
    1. f ( x ) = 1 · b ( x ) - 2
    2. g ( x ) = 1 · b ( x ) - 1
    3. h ( x ) = 1 · b ( x ) + 0
    4. j ( x ) = 1 · b ( x ) + 1
    5. k ( x ) = 1 · b ( x ) + 2
    Use your results to deduce the effect of q .

You should have found that the value of a affects whether the graph curves upwards ( a > 0 ) or curves downwards ( a < 0 ).

You should have also found that the value of q affects the position of the y -intercept.

These different properties are summarised in [link] .

Table summarising general shapes and positions of functions of the form y = a b ( x ) + q .
a > 0 a < 0
q > 0
q < 0

Domain and range

For y = a b ( x ) + q , the function is defined for all real values of x . Therefore, the domain is { x : x R } .

The range of y = a b ( x ) + q is dependent on the sign of a .

If a > 0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q

Therefore, if a > 0 , then the range is { f ( x ) : f ( x ) [ q ; ) } .

If a < 0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q

Therefore, if a < 0 , then the range is { f ( x ) : f ( x ) ( - ; q ] } .

For example, the domain of g ( x ) = 3 · 2 x + 2 is { x : x R } . For the range,

2 x 0 3 · 2 x 0 3 · 2 x + 2 2

Therefore the range is { g ( x ) : g ( x ) [ 2 ; ) } .

Intercepts

For functions of the form, y = a b ( x ) + q , the intercepts with the x and y axis is calculated by setting x = 0 for the y -intercept and by setting y = 0 for the x -intercept.

The y -intercept is calculated as follows:

y = a b ( x ) + q y i n t = a b ( 0 ) + q = a ( 1 ) + q = a + q

For example, the y -intercept of g ( x ) = 3 · 2 x + 2 is given by setting x = 0 to get:

y = 3 · 2 x + 2 y i n t = 3 · 2 0 + 2 = 3 + 2 = 5

The x -intercepts are calculated by setting y = 0 as follows:

y = a b ( x ) + q 0 = a b ( x i n t ) + q a b ( x i n t ) = - q b ( x i n t ) = - q a

Which only has a real solution if either a < 0 or q < 0 . Otherwise, the graph of the function of form y = a b ( x ) + q does not have any x -intercepts.

For example, the x -intercept of g ( x ) = 3 · 2 x + 2 is given by setting y = 0 to get:

y = 3 · 2 x + 2 0 = 3 · 2 x i n t + 2 - 2 = 3 · 2 x i n t 2 x i n t = - 2 3

which has no real solution. Therefore, the graph of g ( x ) = 3 · 2 x + 2 does not have any x -intercepts.

Asymptotes

Functions of the form y = a b ( x ) + q have a single horizontal asymptote. The asymptote can be determined by examining the range.

We have seen that the range is controlled by the value of q. If a > 0 , then the range is { f ( x ) : f ( x ) [ q ; ) } .And if a > 0 , then the range is { f ( x ) : f ( x ) [ q ; ) } .

This shows that the function tends towards the value of q as x . Therefore the horizontal asymptote lies at x = q .

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

In order to sketch graphs of functions of the form, f ( x ) = a b ( x ) + q , we need to determine four characteristics:

  1. domain and range
  2. asymptote
  3. y -intercept
  4. x -intercept

For example, sketch the graph of g ( x ) = 3 · 2 x + 2 . Mark the intercepts.

We have determined the domain to be { x : x R } and the range to be { g ( x ) : g ( x ) [ 2 , ) } .

The y -intercept is y i n t = 5 and there are no x -intercepts.

Graph of g ( x ) = 3 · 2 x + 2 .

Draw the graph of y = - 2 . 3 x + 5 .

  1. The domain is: { x : x R } and the range is: { f ( x ) : f ( x ) ( - ; 5 ] } .
  2. There is one asymptote for functions of this form. This occurs at y = q . So the asymptote for this graph is at y = 5
  3. The y-intercept occurs when x = 0 .
    y = - 2 . 3 x + 5 y = - 2 . 3 0 + 5 y = - 2 ( 1 ) + 5 y int = 7
    So there is one y-intercept at ( 0 , 7 ) .
  4. The x-intercept occurs when y = 0 . Calculating the x-intercept gives:
    y = - 2 . 3 x + 5 0 = - 2 . 3 x + 5 - 5 = - 2 . 3 x 3 x int = 5 2 x int = 0,83
    So there is one x-intercept at ( 0,83 , 0 ) .
  5. Putting all this together gives us the following graph:
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Source:  OpenStax, Siyavula textbooks: grade 10 maths [caps]. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11306/1.4
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