IGCSE Mathematics - Additional (0606) - 2021 Edition

4.07 Absolute value functions

Lesson

The modulus or absolute value of a number can be thought of as **the distance a number is from **$0$0 on a number line. For example, the absolute value of $-3$−3 and $3$3 is $3$3.

Absolute value is represented mathematically by two vertical lines on either side of a value. For example $\left|3\right|$|3| means "*the modulus or absolute value of *$3$3," and equals $3$3, $\left|-9\right|$|−9| means "*the modulus or absolute value of *$-9$−9" and equals $9$9 and the $\left|-x\right|$|−`x`| means "*the modulus or absolute value of* $-x$−`x` " and equals $x$`x`.

Except for $0$0, the absolute value of any real number is the positive value of that number. The absolute value of an expression is the positive value of the number returned by the expression.

Let's consider the basic modulus function $f(x)=|x|$`f`(`x`)=|`x`|. This looks similar to the linear function $f(x)=x$`f`(`x`)=`x` but with absolute value signs. We know that $f(x)=x$`f`(`x`)=`x` represents a straight line through the origin on the coordinate plane. Introducing the absolute value means that negative function values become positive as shown in the table and on the diagram below. Notice that this occurs at the $x$`x` intercept of the graph of $f(x)=x$`f`(`x`)=`x`.

**Table of values:**

$x$x |
$-2$−2 | $-1$−1 | $0$0 | $1$1 | $2$2 |
---|---|---|---|---|---|

$f(x)=\left|x\right|$f(x)=|x| |
$2$2 | $1$1 | $0$0 | $1$1 | $2$2 |

**Graph:**

Let's consider a more complicated modulus function like $y=|5x-15|$`y`=|5`x`−15|. The $x$`x` intercept of the function $y=5x-15$`y`=5`x`−15 is the critical value below which it is negative. This will be the vertex of the absolute value function.

To find the $x$`x` intercept we make $y=0$`y`=0:

$5x-15$5x−15 |
$=$= | $0$0 |

$5x$5x |
$=$= | $15$15 |

$x$x |
$=$= | $3$3 |

Hence, as this graph has a positive gradient function values will be negative for $x<3$`x`<3. For $x>3$`x`>3, the graph of $y=|5x-15|$`y`=|5`x`−15| will be identical the graph of $y=5x-15$`y`=5`x`−15. For the other values, the negative $y$`y` values turn positive.

We can think of modulus functions as being defined by two functions: one for values of $x$`x` that would make the original function positive and another for values of $x$`x` that make the original function negative.

So $y=|5x-15|$`y`=|5`x`−15| can be redefined as:

In the diagram below, the lines corresponding to the two parts of a function definition are shown lightly with the 'composite' absolute value function drawn more heavily.

Other functions of the form $|ax+b|$|`a``x`+`b`| have a shape similar to the one illustrated above. The 'V' shape is typical with the vertex located on the horizontal axis at the point $x$`x` that makes $ax+b=0$`a``x`+`b`=0.

When the coefficient $a$`a` is large, the 'V' shape is narrow. This is because $a$`a` controls the gradients. One side has a gradient of $a$`a` while the other has a gradient of $-a$−`a`. Because of this, these absolute value functions will always be symmetrical.

If a negative sign is placed in front of the absolute value symbol, the effect is to invert the 'V' shape of the graph. All the function values are negative when $y=-|ax+b|$`y`=−|`a``x`+`b`|.

As you can see, our understanding of the features of straight lines are very helpful in sketching modulus functions.

Piecewise definition of $f(x)=|ax+b|$`f`(`x`)=|`a``x`+`b`|

A linear modulus function is really a composite of two functions. It is defined by a rule like the following, where the coefficient $a$`a` can be assumed to be a positive real number.

Particular functions are obtained by substituting appropriate values for the coefficients $a$`a` and $b$`b`.

Consider the function $f\left(x\right)=\left|x\right|$`f`(`x`)=|`x`| that has been graphed. Notice that it opens upwards.

Loading Graph...

What is the gradient of the function for $x>0$

`x`>0?What is the gradient of the function for $x<0$

`x`<0?The graph below shows the graph of $y$

`y`that results from reflecting $f\left(x\right)$`f`(`x`) about the $x$`x`-axis. State the equation of $y$`y`.Loading Graph...Select all the correct statements.

A downward absolute value function goes from decreasing to increasing.

AAn upward absolute value function goes from decreasing to increasing.

BAn upward absolute value function goes from increasing to decreasing.

CA downward absolute value function goes from increasing to decreasing.

DA downward absolute value function goes from decreasing to increasing.

AAn upward absolute value function goes from decreasing to increasing.

BAn upward absolute value function goes from increasing to decreasing.

CA downward absolute value function goes from increasing to decreasing.

D

Consider the graph of function $f\left(x\right)$`f`(`x`).

Loading Graph...

State the coordinate of the vertex.

State the equation of the line of symmetry.

What is the gradient of the function for $x>5$

`x`>5?What is the gradient of the function for $x<5$

`x`<5?Hence, which of the following statements is true?

The graph of $f\left(x\right)$

`f`(`x`) is steeper than the graph of $y=\left|x\right|$`y`=|`x`|.AThe graph of $f\left(x\right)$

`f`(`x`) is not as steep as the graph of $y=\left|x\right|$`y`=|`x`|.BThe graph of $f\left(x\right)$

`f`(`x`) has the same steepness as the graph of $y=\left|x\right|$`y`=|`x`|.CThe graph of $f\left(x\right)$

`f`(`x`) is steeper than the graph of $y=\left|x\right|$`y`=|`x`|.AThe graph of $f\left(x\right)$

`f`(`x`) is not as steep as the graph of $y=\left|x\right|$`y`=|`x`|.BThe graph of $f\left(x\right)$

`f`(`x`) has the same steepness as the graph of $y=\left|x\right|$`y`=|`x`|.C

Consider the function $y=\left|x+2\right|$`y`=|`x`+2|.

Determine the coordinates of the $y$

`y`-intercept.Intercept $=$= $\left(\editable{},\editable{}\right)$(,)

State the coordinate of the vertex.

Vertex $=$= $\left(\editable{},\editable{}\right)$(,)

Draw the graph of the function.

Loading Graph...

We can graph absolute value functions of the form: $y=|ax^2+bx+c|$`y`=|`a``x`2+`b``x`+`c`| by first graphing the quadratic function $y=ax^2+bx+c$`y`=`a``x`2+`b``x`+`c` and then reflecting any parts of the parabola that are below the $x$`x`-axis, by the $x$`x`-axis so that the whole function lies above the $x$`x`-axis. If the entire parabola is already above the $x$`x`-axis, then nothing needs to be reflected.

Graph the function $y=|x^2-4x-12|.$`y`=|`x`2−4`x`−12|.

**Think**: First we should graph the function without the absolute value signs.

**Do**: To graph the function, we should find the $x$`x`-intercepts (when $y=0$`y`=0).

$x^2-4x-12$x2−4x−12 |
$=$= | $0$0 |

$(x-6)(x+2)$(x−6)(x+2) |
$=$= | $0$0 |

$x$x |
$=$= | $-2,6$−2,6 |

We should also find the coordinates of the vertex:

$x$x |
$=$= | $-\frac{b}{2a}$−b2a |

$=$= | $\frac{4}{2}$42 | |

$=$= | $2$2 | |

$y$y |
$=$= | $2^2-4\times2-12$22−4×2−12 |

$=$= | $4-8-12$4−8−12 | |

$=$= | $-16$−16 |

Therefore the coordinates of the vertex are $(2,-16)$(2,−16).

So now we are ready to graph the parabola without the absolute value signs:

We now need to reflect the part of the parabola that is below the $x$`x`-axis, by the $x$`x`-axis, so that the absolute value function is above the $x$`x`-axis:

Note that the vertex has been reflected to the point $(2,16),$(2,16),and the $y$`y`-intercept has been reflected to $y=12$`y`=12 from $y=-12$`y`=−12.

So the final graph of the function $y=|x^2-4x-12|$`y`=|`x`2−4`x`−12| is:

Understand the relationship between y = f(x) and y = |f(x)|, where f(x) may be linear, quadratic or trigonometric.

Know the conditions for f(x) = 0 to have two real roots, two equal roots, no real roots. Know the related conditions for a given line to intersect a given curve, be a tangent to a given curve, not intersect a given curve.

Find the solution set for quadratic inequalities.