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## Functions of the form $y=1x$

Functions of the general form $y=ax+q$ are called hyperbolic functions.

### Example 1: Plotting a hyperbolic function

#### Question

$y=h(x)=1x$(1)

Complete the following table for $h(x)=1x$ and plot the points on a system of axes.

Table 1
 x −3 −2 −1 $-12$ $-14$ 0 $14$ $12$ 1 2 3 $h(x)$ $-13$ $-2$ $-2$ $-2$ $-2$ $-2$ $-2$ $-2$ $-2$ $-2$ $-2$
1. Join the points with smooth curves.

2. What happens if $x=0$?

3. Explain why the graph consists of two separate curves.

4. What happens to $h(x)$ as the value of $x$ becomes very small or very large?

5. The domain of $h(x)$ is ${x:x∈ℝ,x≠0}$. Determine the range.

6. About which two lines is the graph symmetrical?

##### Substitute values into the equation
$h(x)=1xh(-3)=1-3=-13h(-2)=1-2=-12h(-1)=1-1=-1h(-12)=1-12=-2h(-14)=1-14=-4h(0)=10=undefined$(2)
$h(14)=114=4h(12)=112=2h(1)=11=1h(2)=12=12h(3)=13=13$(3)
Table 2
 x −3 −2 −1 $-12$ $-14$ 0 $14$ $12$ 1 2 3 $h(x)$ $-13$ $-12$ −1 −2 −4 undefined 4 2 1 $12$ $13$
##### Plot the points and join with two smooth curves

From the table we get the following points: $(-3;-13)$, $(-2;-12)$, $(-1;-1)$, $(-12;-2)$, $(-14;-4)$, $(14;4)$, $(12;2)$, $(1;1)$, $(2;12)$, $(3;13)$.

For $x=0$ the function $h$ is undefined. This is called a discontinuity at $x=0$.

$y=h(x)=1x$ therefore we can write that $x×y=1$. Since the product of two positive numbers and the product of two negative numbers can be equal to 1, the graph lies in the first and third quadrants.

##### Determine the asymptotes

As the value of $x$ gets larger, the value of $h(x)$ gets closer to, but does not equal 0. This is a horizontal asymptote, the line $y=0$. The same happens in the third quadrant; as $x$ gets smaller, $h(x)$ also approaches the negative $x$-axis asymptotically.

We also notice that there is a vertical asymptote, the line $x=0$; as $x$ gets closer to 0, $h(x)$ approaches the $y$-axis asymptotically.

##### Determine the range

Domain: ${x:x∈ℝ,x≠0}$

From the graph, we see that $y$ is defined for all values except 0.

Range: ${y:y∈ℝ,y≠0}$

##### Determine the lines of symmetry

The graph of $h(x)$ has two axes of symmetry: the lines $y=x$ and $y=-x$. About these two lines, one half of the hyperbola is a mirror image of the other half.

## Functions of the form $y=ax+q$

### Investigation 1: The effects of $a$ and $q$ on a hyperbola

On the same set of axes, plot the following graphs:

1. $y1=1x-2$

2. $y2=1x-1$

3. $y3=1x$

4. $y4=1x+1$

5. $y5=1x+2$

Use your results to deduce the effect of $q$.

On the same set of axes, plot the following graphs:

1. $y6=-2x$

2. $y7=-1x$

3. $y8=1x$

4. $y9=2x$

Use your results to deduce the effect of $a$.

The effect of $q$

The effect of $q$ is called a vertical shift because all points are moved the same distance in the same direction (it slides the entire graph up or down).

• For $q>0$, the graph of $f(x)$ is shifted vertically upwards by $q$ units.

• For $q<0$, the graph of $f(x)$ is shifted vertically downwards by $q$ units.

The horizontal asymptote is the line $y=q$ and the vertical asymptote is always the $y$-axis, the line $x=0$.

The effect of $a$

The sign of $a$ determines the shape of the graph.

• If $a>0$, the graph of $f(x)$ lies in the first and third quadrants.

For $a>1$, the graph of $f(x)$ will be further away from the axes than $y=1x$.

For $0, as a tends to 0, the graph moves closer to the axes than $y=1x$.

• If $a<0$, the graph of $f(x)$ lies in the second and fourth quadrants.

For $a<-1$, the graph of $f(x)$ will be further away from the axes than $y=-1x$.

For $-1, as a tends to 0, the graph moves closer to the axes than $y=-1x$.

## Discovering the characteristics

The standard form of a hyperbola is the equation $y=ax+q$.

### Domain and range

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

We see that $y=ax+q$ can be re-written as:

$y=ax+qy-q=axIfx≠0then:(y-q)x=ax=ay-q$(4)

This shows that the function is undefined only at $y=q$.

Therefore the range is ${f(x):f(x)∈ℝ,f(x)≠q}$.

#### Example 2: Domain and range of a hyperbola

##### Question

If $g(x)=2x+2$, determine the domain and range of the function.

###### Determine the domain

The domain is ${x:x∈ℝ,x≠0}$ because $g(x)$ is undefined only at $x=0$.

###### Determine the range

We see that $g(x)$ is undefined only at $y=2$. Therefore the range is ${g(x):g(x)∈ℝ,g(x)≠2}$.

### Intercepts

The $y$-intercept:

Every point on the $y$-axis has an $x$-coordinate of 0, therefore to calculate the $y$-intercept, let $x=0$.

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

$y=2x+2y=20+2$(5)

which is undefined, therefore there is no $y$-intercept.

The $x$-intercept:

Every point on the $x$-axis has a $y$-coordinate of 0, therefore to calculate the $x$-intercept, let $y=0$.

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

$y=2x+20=2x+22x=-2x=2-2=-1$(6)

This gives the point $(-1;0)$.

### Asymptotes

There are two asymptotes for functions of the form $y=ax+q$.

The horizontal asymptote is the line $y=q$ and the vertical asymptote is always the $y$-axis, the line $x=0$.

### Axes of symmetry

There are two lines about which a hyperbola is symmetrical: $y=x+q$ and $y=-x+q$.

## Sketching graphs of the form $y=ax+q$

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

1. sign of $a$

2. $y$-intercept

3. $x$-intercept

4. asymptotes

### Example 3: Sketching a hyperbola

#### Question

Sketch the graph of $g(x)=2x+2$. Mark the intercepts and the asymptotes.

##### Examine the standard form of the equation

We notice that $a>0$ therefore the graph of $g(x)$ lies in the first and third quadrant.

##### Calculate the intercepts

For the $y$-intercept, let $x=0$:

$g(x)=2x+2g(0)=20+2$(7)

This is undefined, therefore there is no $y$-intercept.

For the $x$-intercept, let $y=0$:

$g(x)=2x+20=2x+22x=-2∴x=-1$(8)

This gives the point $(-1;0)$.

##### Determine the asymptotes

The horizontal asymptote is the line $y=2$. The vertical asymptote is the line $x=0$.

##### Sketch the graph

Domain: ${x:x∈ℝ,x≠0}$.

Range: ${y:y∈ℝ,y≠2}$.

### Example 4: Sketching a hyperbola

#### Question

Sketch the graph of $y=-4x+7$.

##### Examine the standard form of the equation

We see that $a<0$ therefore the graph lies in the second and fourth quadrants.

##### Calculate the intercepts

For the $y$-intercept, let $x=0$:

$y=-4x+7=-40+7$(9)

This is undefined, therefore there is no $y$-intercept.

For the $x$-intercept, let $y=0$:

$y=-4x+70=-4x+7-4x=-7∴x=47$(10)

This gives the point $47;0$.

##### Determine the asymptotes

The horizontal asymptote is the line $y=7$. The vertical asymptote is the line $x=0$.

##### Sketch the graph

Domain: ${x:x∈ℝ,x≠0}$

Range: ${y:y∈ℝ,y≠7}$

Axis of symmetry: $y=x+7$ and $y=-x+7$

### Exercise 1:

Draw the graph of $xy=-6$.

1. Does the point $(-2;3)$ lie on the graph? Give a reason for your answer.

2. If the $x$-value of a point on the drawn graph is 0,25 what is the corresponding $y$-value?

3. What happens to the $y$-values as the $x$-values become very large?

4. Give the equations of the asymptotes.

5. With the line $y=-x$ as line of symmetry, what is the point symmetrical to $(-2;3)$?

Draw the graph of $h(x)=8x$.

1. How would the graph $g(x)=8x+3$ compare with that of $h(x)=8x$? Explain your answer fully.

2. Draw the graph of $y=8x+3$ on the same set of axes, showing asymptotes, axes of symmetry and the coordinates of one point on the graph.

To-do.

To-do.