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CanadaON
Grade 11

Mixed Graphs II (linear, quad, cubic, 1/x)

Lesson

Reviewing some basic graph types

Let's first recall the basic graphs as shown here in the diagram. The function's description is given below each graph. For example the linear function $y=mx+b$y=mx+b is shown on the left and the hyperbola $y=\frac{1}{x}$y=1x is shown second from right. 

Two of the graphs are shown with some generalising coefficients - the line $y=mx+b$y=mx+b and the exponential curve  $y=a^x$y=ax. Thus we could site particular examples of these graphs such as $y=2x+5$y=2x+5, $y=7x$y=7x and $y=3^x$y=3x etc.

However, all of these basic types can be generalised further by what are known as transformations.

Translations

One common transformation is known as a translation.  This is a horizontal and/or vertical shift in the position of the curve relative to the cartesian axes. Translations are easy to recognise.

If we replace $x$x with $\left(x-h\right)$(xh), the basic curve is moved $h$h units to the right. If we replace $x$x with $\left(x+h\right)$(x+h), the basic curve is moved $h$h units to the left. If we replace $y$y with $\left(y-k\right)$(yk), the basic curve is moved $k$k units up. If we replace $y$y with $\left(y+k\right)$(y+k), the basic curve is moved $k$k units down. 

So suppose we take the hyperbola $y=\frac{1}{x}$y=1x and translate it $3$3 units to the right and $5$5 units down. The new function becomes $y+5=\frac{1}{x-3}$y+5=1x3, or expressed explicitly $y=\frac{1}{x-3}-5$y=1x35. The vertical and horizontal asymptotes are respectively $x=3$x=3 and $y=-5$y=5.

If we take the parabola $y=x^2$y=x2 and shift it $1$1 unit to the left and $5$5 units up, the new function becomes $y-5=\left(x-1\right)^2$y5=(x1)2 . Again, we can express this explicitly as $y=\left(x-1\right)^2+5$y=(x1)2+5

If we take the function $y=x^3$y=x3 and translate it $2$2 units to the right and $1$1 unit up, the new function is given by $y-1=\left(x-2\right)^3$y1=(x2)3, expressible as $y=\left(x-2\right)^3+1$y=(x2)3+1

Dilations and reflections

Another type of transformation is commonly referred to as a dilation. This is when a curve is stretched or compressed by some factor other than $1$1 in the function's equation.

For example, the difference between $y=x^2$y=x2 to $y=3x^2$y=3x2 is the dilation factor $3$3. Every function value in $y=3x^2$y=3x2 is $3$3 times the associated function value in $y=x^2$y=x2. This means that the curve becomes steeper.

Again, every function value of the function $y=\frac{2}{x}$y=2x is double the associated function value of $y=\frac{1}{x}$y=1x.

Every function value of $y=3\left(2^x\right)$y=3(2x) is $3$3 times that of $y=2^x$y=2x, and every function value of $y=\frac{1}{2}x^3$y=12x3 is one-halve that of $y=x^3$y=x3.

We can also reflect a curve across the $x$x - axis by simply multiplying its function expression by $-1$1.

Every function value of $y=-x^2$y=x2 is the negative of those of $y=x^2$y=x2, and the curve becomes an upside-down parabola with the same vertex as $y=x^2$y=x2. Again, every function value of $y=-\frac{3}{x}$y=3x is the reflected image of across the x -axis of $y=\frac{3}{x}$y=3x.  

 

Combining Translations and dilations

When we combine translations and dilations we can reposition and distort the basic curve. 

For example, the curve of $y=\frac{5}{x-2}+3$y=5x2+3 is constructed as follows. Start with the curve given by the basic function $y=\frac{1}{x}$y=1x and stretch each ordinate by a factor of $5$5. Then shift it $2$2 units to the right  and $3$3 units up, so that the centre becomes $\left(2,3\right)$(2,3)  with the curve's asymptotes as  $x=2$x=2 and $y=3$y=3.

The final graph is shown in red here, along with the graph of $y=\frac{1}{x-2}+3$y=1x2+3. Note how changing from $y=\frac{1}{x-2}+3$y=1x2+3 to $y=\frac{5}{x-2}+3$y=5x2+3 pushes the hyperbola out diagonally. 

Finding $x$x and $y$y intercepts    

The $y$y intercept of any function are found by putting $x=0$x=0 into the equation. The $x$x intercept(s) are found similarly by putting $x=0$x=0

For example, for the curve given by $y=\frac{1}{x-3}-5$y=1x35, putting $x=0$x=0 reveals $y=\frac{1}{0-3}-5=-5\frac{1}{3}$y=1035=513

The x intercept is found by putting $y=0$y=0, so that $\frac{1}{x-3}-5=0$1x35=0. This means that $\frac{1}{x-3}=5$1x3=5 and inverting $x-3=\frac{1}{5}$x3=15 so that $x=3\frac{1}{5}$x=315.

 

 

 

Worked Examples

Question 1

Consider the equation $y=-2x$y=2x.

  1. Find the $y$y-value of the $y$y-intercept of the line.

  2. Find the $x$x-value of the $x$x-intercept of the line.

  3. Find the value of $y$y when $x=2$x=2.

  4. Plot the equation of the line below.

    Loading Graph...

Question 2

Consider the function $y=-\frac{1}{2}x^2$y=12x2

  1. Complete the following table of values.

    $x$x $-2$2 $-1$1 $0$0 $1$1 $2$2
    $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. Plot the graph.

    Loading Graph...

Question 3

Consider the cubic function $y=4x^3-3$y=4x33

  1. Is the cubic increasing or decreasing from left to right?

    Increasing

    A

    Decreasing

    B
  2. Is the cubic more or less steep than the function $y=x^3$y=x3 ?

    More steep

    A

    Less steep

    B
  3. What are the coordinates of the point of inflection of the function?

    Inflection ($\editable{}$, $\editable{}$)

  4. Plot the graph $y=4x^3-3$y=4x33

    Loading Graph...

Question 4

Consider the function $y=-\frac{1}{4x}$y=14x

  1. Complete the following table of values.

    $x$x $-3$3 $-2$2 $-1$1 $1$1 $2$2 $3$3
    $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. Sketch the graph.

    Loading Graph...

  3. In which quadrants does the graph lie?

    $1$1

    A

    $2$2

    B

    $3$3

    C

    $4$4

    D

 

Outcomes

11U.A.1.8

Determine, through investigation using technology, the roles of the parameters a, k, d, and c in functions of the form y = a f (k(x – d)) + c, and describe these roles in terms of transformations on the graphs of f(x) = x, f(x) = x^2 , f(x) = sqrt(x) and f(x) = 1/x

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