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1.01 Translations and dilations of known functions

Lesson

Function types

A family of functions is a set of functions whose equations have a similar form. The parent function of the family is the equation in the family with the simplest form. Some common function types are shown in the table below.

Linear $f(x)=x$f(x)=x

The linear function has the form $y=mx+c$y=mx+c. As its name suggests, it is drawn as a straight line that intersects the $y$y- axis at$\left(0,c\right)$(0,c) and increases or decreases according to the value of the coefficient m, which represents the gradient.

Quadratic $f(x)=x^2$f(x)=x2

The quadratic function, or parabola $y=x^2$y=x2, takes $x$x-values and squares them. Non-zero numbers when squared become positive, and that explains the symmetry of the graph.

Moreover, as the absolute value of $x$x becomes large, $x^2$x2 becomes very large and that's why the curve becomes steep on both sides of the axis of symmetry. In fact, all even powered polynomials act in this general way.

Cubic $f(x)=x^3$f(x)=x3

The cubic function $y=x^3$y=x3 grows even quicker than the parabola, but negative numbers when cubed remain negative. This explains the curve's shape. In this basic form, it possesses $180^\circ$180° rotational symmetry about the origin. Odd powered polynomials act in this general way, with large positive $y$y-values for positive $x$x-values and large negative $y$y-values for negative $x$x-values.

Semicircle $f(x)=\sqrt{r^2-x^2}$f(x)=r2x2

Although a circle is a relation, a semicircle, where we restrict $y$y-values to leave us with either the top or bottom half of the circle, is a function. The equation $f(x)=\sqrt{r^2-x^2}$f(x)=r2x2 defines a semicircle with its centre at the origin and a radius $r$r.

Absolute value $f(x)=\left|x\right|$f(x)=|x|

The absolute value function given as $y=\left|x\right|$y=|x| leaves positive values of the line $y=x$y=x unchanged but changes the negative values of the same line to positive values. Negative values are thus reflected across the $x$x-axis, and this is why the function exhibits its angular shape.

Reciprocal (hyperbola) $f(x)=\frac{1}{x}$f(x)=1x

The reciprocal function, or hyperbola, possesses positive function values of $y=\frac{1}{x}$y=1x that are large when $x$x is small and positive, and small when $x$x is large and positive. In other words, the function takes small values and turns them into large values and vice-versa. As the denominator of a fraction cannot be zero, and it is possible to rearrange this equation to have either $x$x or $y$y as the denominator, a hyperbola has a vertical and a horizontal asymptote. In its basic form, the hyperbola is another function that possesses $180^\circ$180° rotational symmetry about the origin.

Exponential $f(x)=b^x$f(x)=bx, $b>1$b>1

The exponential function$y=a^x$y=ax, $a>1$a>1 has the independent variable in the exponent. The function values are always positive and increase rapidly as $x$x continues to increase, giving it the characteristic shape as the $y$y-values exponentially increase.

Logarithmic $f(x)=\log_bx$f(x)=logbx, $b>1$b>1

The logarithmic function is the inverse of the exponential function and as such is only defined for values $x>0$x>0. As $x$x approaches $0$0, the $y$y-values steadily decrease, forming a vertical asymptote. The $y$y-values always increase for increasing $x$x, but for larger values of $x$x the rate of this increase is slower.

Of course, we are expected to sketch and interpret all kinds of transformed versions of these basic functions. We will now focus on learning the general patterns and rules that apply to these transformations.

Reflections

Reflections were covered in detail in the Year 11 course, and are summarised below:

  • For $f(x)$f(x), the transformed function $g(x)=-f(x)$g(x)=f(x) is $f(x)$f(x) reflected over the $x$x-axis.
  • For $f(x)$f(x), the transformed function $g(x)=f(-x)$g(x)=f(x) is $f(x)$f(x) reflected over the $y$y-axis.

 

 

Practice question

Question 1

Consider a graph of $y=4^x$y=4x.

  1. How could the graph of $y=4^{-x}$y=4x be obtained from the graph of $y=4^x$y=4x?

    by making it steeper

    A

    through a reflection across the $x$x-axis

    B

    through a reflection across the $y$y-axis

    C

    through a vertical translation

    D
  2. Given the graph of $y=4^x$y=4x, graph $y=4^{-x}$y=4x on the same set of axes.

    Loading Graph...

Translations

Any of the basic functions can be moved up or down $c$c units by simply adding $c$c. For example, the graph of the function $y=x^2+3$y=x2+3 is essentially the same as $y=x^2$y=x2 but moved up (translated vertically) $3$3 units. The same idea applies to any function.

Any of the graphs can be moved sideways $b$b units by simply replacing $x$x with $\left(x-b\right)$(xb). So, for example, the cubic graph $y=\left(x-3\right)^3$y=(x3)3 is essentially the same as $y=x^3$y=x3 but moved to the right (translated horizontally to the right) by $3$3 units. Again, the same idea applies to any function.

The image below summarises the four translations possible, using a hyperbola as the basic function.

 

Note that we add $c$c to move a function up $c$c units in the positive direction, but need to subtract $b$b to move a graph $b$b units in the positive direction when we are looking at a sideways shift. In actual fact, the process can be consolidated to just one technique of subtracting, as long as you make this change to the variable in question. 

Take $y=\left|x\right|$y=|x| for example. To move that $3$3 units to the right and $4$4 units up, we could write the resulting function as $y=\left|x-3\right|+4$y=|x3|+4. We could have written it as $y-4=\left|x-3\right|$y4=|x3| and produced the same translation.

This technique is used when translating circles. Because of the squared function on $y$y, if we subtract a value of $c$c from the $c$c variable it creates a shift upwards of $c$c units. For example, consider the circle centred on the origin with radius $2$2 which has the equation $x^2+y^2=4$x2+y2=4.  Suppose we replace the $x$x with $\left(x-15\right)$(x15) and the $y$y with $\left(y-9\right)$(y9) so that our new equation becomes $\left(x-15\right)^2+\left(y-9\right)^2=4$(x15)2+(y9)2=4.

The centres of the two circles are depicted in the diagram below. 

Summary: Translations

$g(x)=f(x)+c$g(x)=f(x)+c is the graph of $f(x)$f(x) translated vertically
  • If $c>0$c>0, then translated $c$c units up
  • If $c<0$c<0, then translated $c$c units down
$g(x)=f(x+b)$g(x)=f(x+b) is the graph of $f(x)$f(x) translated horizontally
  • If $b>0$b>0, then translated $b$b units left
  • If $b<0$b<0, then translated $b$b units right

Practice questions

Question 2

This is a graph of $y=\frac{1}{x}$y=1x.

Loading Graph...
A cartesian plane is shown with both the x-axis and y-axis ranging from -10 to 10. A graph of a hyperbola with a function $y=\frac{1}{x}$y=1x is plotted. 
  1. How do we shift the graph of $y=\frac{1}{x}$y=1x to get the graph of $y=\frac{1}{x+2}$y=1x+2?

    Shift the graph to the right by $2$2 units..

    A

    Shift the graph upwards by $2$2 units.

    B

    Shift the graph to the left by $2$2 units.

    C

    Shift the graph downwards by $2$2 units.

    D
  2. Hence plot $y=\frac{1}{x+2}$y=1x+2 on the same graph as $y=\frac{1}{x}$y=1x.

     

    Loading Graph...
    A cartesian plane is shown with both the x-axis and y-axis ranging from -10 to 10. A hyperbola is drawn with a function y = 1/x.

Question 3

Consider the graph of $y=x^2-4$y=x24.

Loading Graph...

A parabola that opens upward on a Cartesian plane with its vertex located at the coordinates (0, -4). The coordinates of the vertex are not explicitly labeled or given.
  1. What would be the new equation if the graph was translated upwards by $9$9 units?

    $y=\left(x-9\right)^2-4$y=(x9)24

    A

    $y=x^2-13$y=x213

    B

    $y=x^2+5$y=x2+5

    C

    $y=\left(x+9\right)^2-4$y=(x+9)24

    D
  2. What would be the new equation if the graph was translated to the right by $4$4 units?

    $y=\left(x-4\right)^2-4$y=(x4)24

    A

    $y=\left(x+4\right)^2-4$y=(x+4)24

    B

    $y=x^2-8$y=x28

    C

    $y=x^2$y=x2

    D

Question 4

We want to graph $y=\left|x\right|+2$y=|x|+2.

  1. What is the domain of the function?

    $-22<x<2

    A

    $x\ge0$x0

    B

    $-\infty<x<

    C

    $x\le0$x0

    D
  2. What is the range of the function?

    $y\le0$y0

    A

    $-\infty<y<

    B

    $y\ge0$y0

    C

    $y\ge2$y2

    D
  3. Now we can graph $y=\left|x\right|+2$y=|x|+2.

    Loading Graph...

Dilations

Another type of transformation is commonly referred to as dilation. This is when a curve is stretched or compressed by some factor.

For example, the difference between $y=x^2$y=x2 and $y=3x^2$y=3x2 is the vertical dilation factor $3$3. For the same $x$x-value, every $y$y-value in $y=3x^2$y=3x2 is $3$3 times the $y$y-value in $y=x^2$y=x2. This means that the curve $y=3x^2$y=3x2 becomes steeper at a faster rate. Similarly, every function value of the function $y=\frac{1}{2}\log_2(x)$y=12log2(x) is half the associated function value of $y=\log_2(x)$y=log2(x), so the curve will be compressed. Since the multiplication is effectively being applied to the whole function $y$y, we reason that the curves are changing only in relation to their position on the $x$x-axis, hence this type of dilation is considered to be vertical dilation.

The final transformation we need to consider occurs when we multiply the $x$x values within a function by a constant. This is a transformation by a factor of $a$a in the form $y=f(ax)$y=f(ax). We saw above that multiplying the $y$y values of the function resulted in a dilation away from the $x$x-axis. So, we could make a conjecture that this $y=f(ax)$y=f(ax) transformation will result in a dilation away from the $y$y-axis. 

In the diagram below, we have sketched $f(x)=x^3-4x$f(x)=x34x in green, and two transformations: $g(x)=f(2x)=8x^3-8x$g(x)=f(2x)=8x38x in blue and $h(x)=f(\frac{1}{2}x)=\frac{1}{8}x^3-2x$h(x)=f(12x)=18x32x in red.

 

As you can see, $g(x)=f(2x)$g(x)=f(2x) has been compressed towards the $y$y-axis, while $h(x)=f(\frac{1}{2}x)$h(x)=f(12x)has been stretched away from the $y$y-axis. By looking at the $x$x-intercepts we can see that the factor of the compression and stretching matches the coefficient of $x$x: $g(x)$g(x) values are twice as close to the $y$y-axis, while $h(x)$h(x) values are twice as far. Of course, the $x$x-intercept at $x=0$x=0 hasn't changed, because multiplying zero by any sized constant won't change it!

The diagram also very clearly shows that while the $y$y values of the curve have been shifted relative to their original position, the maximum and minimum values of the curve remain the same. Just like the other transformations we have seen, only one aspect of the curve has been changed by applying this transformation.

Summary: Dilations
$g(x)=kf(x)$g(x)=kf(x) is the graph of $f(x)$f(x) dilated vertically by a factor of $k$k
  • If $k>1$k>1, then vertical dilation (stretch) away from the $x$x-axis
  • If$00<k<1, then vertical dilation (compression) toward the $x$x-axis
$g(x)=f(ax)$g(x)=f(ax) is the graph of $f(x)$f(x) dilated horizontally by a factor of $\frac{1}{a}$1a
  • If $a>1$a>1, then horizontal dilation (compression) toward the $y$y-axis
  • If $00<a<1, then horizontal dilation (stretch) away from the $y$y-axis

 

Did you notice?

Transformations inside the function $f(x)$f(x), such as $f(x+3)$f(x+3) and $f(3x)$f(3x), are horizontal.

Transformations outside the function $f(x)$f(x), such as $f(x)+3$f(x)+3 and $3f(x)$3f(x), are vertical.

Practice questions

Question 5

Consider the original graph $y=3^x$y=3x. The function values of the graph are multiplied by $2$2 to form a new graph.

  1. For each point on the original graph, find the point on the new graph.

    Point on original graph Point on new graph
    $\left(-1,\frac{1}{3}\right)$(1,13) $\left(-1,\editable{}\right)$(1,)
    $\left(0,1\right)$(0,1) $\left(0,\editable{}\right)$(0,)
    $\left(1,3\right)$(1,3) $\left(1,\editable{}\right)$(1,)
    $\left(2,9\right)$(2,9) $\left(2,\editable{}\right)$(2,)
  2. What is the equation of the new graph?

  3. Which of the following shows the correct graphs of $y=3^x$y=3x and $y=2\left(3^x\right)$y=2(3x) on the same cartesian plane?

    Loading Graph...

    A

    Loading Graph...

    B

    Loading Graph...

    C
  4. Select the two correct statements.

    For negative $x$x values, $2\left(3^x\right)$2(3x) is above $3^x$3x.

    A

    For positive $x$x values, $2\left(3^x\right)$2(3x) is below $3^x$3x.

    B

    For negative $x$x values, $2\left(3^x\right)$2(3x) is below $3^x$3x.

    C

    For positive $x$x values, $2\left(3^x\right)$2(3x) is above $3^x$3x.

    D

Question 6

How does the graph of $y=\frac{1}{2}x^3$y=12x3 differ to the graph of $y=x^3$y=x3?

  1. One is a reflection of the other across the $y$y-axis

    A

    $y$y increases more rapidly on $y=\frac{1}{2}x^3$y=12x3 than on $y=x^3$y=x3

    B

    $y=\frac{1}{2}x^3$y=12x3 is a horizontal shift of $y=x^3$y=x3

    C

    $y$y increases more slowly on $y=\frac{1}{2}x^3$y=12x3 than on $y=x^3$y=x3

    D

Question 7

Consider the functions $y=2x^4$y=2x4 and $y=\frac{1}{2}x^4$y=12x4.

  1. The graph of $y=x^4$y=x4 has been provided on the coordinate axes below. On the same set of axes, graph $y=2x^4$y=2x4. (You need to use all five points to graph the function)

    Loading Graph...

  2. How would the graph of $y=\frac{1}{2}x^4$y=12x4 differ from the graph of $y=2x^4$y=2x4?

    $y=2x^4$y=2x4 is stretched horizontally and would appear wider than $y=\frac{1}{2}x^4$y=12x4

    A

    $y=2x^4$y=2x4 is compressed horizontally and would appear narrower than $y=\frac{1}{2}x^4$y=12x4

    B

    $y=2x^4$y=2x4 is a vertical translation of $y=\frac{1}{2}x^4$y=12x4

    C

Outcomes

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