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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:
Got questions? Get instant answers now!

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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Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
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lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
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12, 17, 22.... 25th term
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12, 17, 22.... 25th term
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AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
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salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
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can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I'm not sure why it wrote it the other way
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I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
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Commplementary angles
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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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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