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# Hyperbolic functions

## Functions of the Form y=ax+qy=ax+q

Functions of the form y=ax+qy=ax+q are known as hyperbolic functions. The general form of the graph of this function is shown in Figure 1.

### Investigation : Functions of the Form y=ax+qy=ax+q

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

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

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

These different properties are summarised in Table 1. The axes of symmetry for each graph are shown as a dashed line.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

### Domain and Range

For y=ax+qy=ax+q, the function is undefined for x=0x=0. The domain is therefore {x:xR,x0}{x:xR,x0}.

We see that y=ax+qy=ax+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 y = a x + q y - q = a x If x 0 then : ( y - q ) x = a x = a y - q
(1)

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

For example, the domain of g(x)=2x+2g(x)=2x+2 is {x:xR,x0}{x:xR,x0} because g(x)g(x) is undefined at x=0x=0.

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

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

### Intercepts

For functions of the form, y=ax+qy=ax+q, the intercepts with the xx and yy axis is calculated by setting x=0x=0 for the yy-intercept and by setting y=0y=0 for the xx-intercept.

The yy-intercept is calculated as follows:

y = a x + q y i n t = a 0 + q y = a x + q y i n t = a 0 + q
(3)

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

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

y = 2 x + 2 y i n t = 2 0 + 2 y = 2 x + 2 y i n t = 2 0 + 2
(4)

which is undefined.

The xx-intercepts are calculated by setting y=0y=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 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
(5)

For example, the xx-intercept of g(x)=2x+2g(x)=2x+2 is given by setting x=0x=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 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
(6)

### Asymptotes

There are two asymptotes for functions of the form y=ax+qy=ax+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=0x=0 and for y=qy=q. Therefore the asymptotes are x=0x=0 and y=qy=q.

For example, the domain of g(x)=2x+2g(x)=2x+2 is {x:xR,x0}{x:xR,x0} because g(x)g(x) is undefined at x=0x=0. We also see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}.

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

### Sketching Graphs of the Form f(x)=ax+qf(x)=ax+q

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

1. domain and range
2. asymptotes
3. yy-intercept
4. xx-intercept

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

We have determined the domain to be {x:xR,x0}{x:xR,x0} and the range to be {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}. Therefore the asymptotes are at x=0x=0 and y=2y=2.

There is no yy-intercept and the xx-intercept is xint=-1xint=-1.

#### Exercise 1: Drawing a hyperbola

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

#### Graphs

1. Using graph (grid) paper, draw the graph of xy=-6xy=-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 xx-value of a point on the drawn graph is 0,25, what is the corresponding yy-value ?
4. What happens to the yy-values as the xx-values become very large ?
5. With the line y=-xy=-x as line of symmetry, what is the point symmetrical to (-2; 3) ?
2. Draw the graph of xy=8xy=8.
1. How would the graph y=83+3y=83+3 compare with that of xy=8xy=8? Explain your answer fully.
2. Draw the graph of y=83+3y=83+3 on the same set of axes.

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