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Functions of the form y = a x + q

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

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

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

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

You should have found that the value of a affects whether the graph is located in the first and third quadrants of Cartesian plane.

You should have also found that the value of q affects whether the graph lies above the x -axis ( q > 0 ) or below the x -axis ( q < 0 ).

These different properties are summarised in [link] . The axes of symmetry for each graph are shown as a dashed line.

Table summarising general shapes and positions of functions of the form y = a x + q . The axes of symmetry are shown as dashed lines.
a > 0 a < 0
q > 0
q < 0

Domain and range

For y = a x + q , the function is undefined for x = 0 . The domain is therefore { x : x R , x 0 } .

We see that y = a x + q can be re-written as:

y = a x + q y - q = a x If x 0 then : ( y - q ) x = a x = a y - q

This shows that the function is undefined at y = q . Therefore the range of f ( x ) = a x + q is { f ( x ) : f ( x ) ( - ; q ) ( q ; ) } .

For example, the domain of g ( x ) = 2 x + 2 is { x : x R , x 0 } because g ( x ) is undefined at x = 0 .

y = 2 x + 2 ( y - 2 ) = 2 x If x 0 then : x ( y - 2 ) = 2 x = 2 y - 2

We see that g ( x ) is undefined at y = 2 . Therefore the range is { g ( x ) : g ( x ) ( - ; 2 ) ( 2 ; ) } .

Intercepts

For functions of the form, y = a 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 x + q y i n t = a 0 + q

which is undefined because we are dividing by 0. Therefore there is no y -intercept.

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

y = 2 x + 2 y i n t = 2 0 + 2

which is undefined.

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

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

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

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

Asymptotes

There are two asymptotes for functions of the form y = a x + q . Just a reminder, an asymptote is a straight or curved line, which the graph of a function will approach, but never touch. They are determined by examining the domain and range.

We saw that the function was undefined at x = 0 and for y = q . Therefore the asymptotes are x = 0 and y = q .

For example, the domain of g ( x ) = 2 x + 2 is { x : x R , x 0 } because g ( x ) is undefined at x = 0 . We also see that g ( x ) is undefined at y = 2 . Therefore the range is { g ( x ) : g ( x ) ( - ; 2 ) ( 2 ; ) } .

From this we deduce that the asymptotes are at x = 0 and y = 2 .

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

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

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

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

We have determined the domain to be { x : x R , x 0 } and the range to be { g ( x ) : g ( x ) ( - ; 2 ) ( 2 ; ) } . Therefore the asymptotes are at x = 0 and y = 2 .

There is no y -intercept and the x -intercept is x i n t = - 1 .

Graph of g ( x ) = 2 x + 2 .

Draw the graph of y = - 4 x + 7 .

  1. The domain is: { x : x R , x 0 } and the range is: { f ( x ) : f ( x ) ( - ; 7 ) ( 7 ; ) } .
  2. We look at the domain and range to determine where the asymptotes lie. From the domain we see that the function is undefined when x = 0 , so there is one asymptote at x = 0 . The other asymptote is found from the range. The function is undefined at y = q and so the second asymptote is at y = 7
  3. There is no y-intercept for graphs of this form.
  4. The x-intercept occurs when y = 0 . Calculating the x-intercept gives:
    y = - 4 x + 7 0 = - 4 x + 7 - 7 = - 4 x x int = 4 7
    So there is one x-intercept at ( 4 7 , 0 ) .
  5. Putting all this together gives us the following graph:
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Graphs

  1. Using graph (grid) paper, draw the graph of x y = - 6 .
    1. Does the point (-2; 3) lie on the graph ? Give a reason for your answer.
    2. Why is the point (-2; -3) not on the graph ?
    3. If the x -value of a point on the drawn graph is 0,25, what is the corresponding y -value ?
    4. What happens to the y -values as the x -values become very large ?
    5. With the line y = - x as line of symmetry, what is the point symmetrical to (-2; 3) ?
  2. Draw the graph of x y = 8 .
    1. How would the graph y = 8 3 + 3 compare with that of x y = 8 ? Explain your answer fully.
    2. Draw the graph of y = 8 3 + 3 on the same set of axes.

Questions & Answers

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Maira Reply
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Ali
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
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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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