ALGEBRA · HL
Absolute Value
The size of a number — always positive.
Section 1 of 5
What is absolute value?
Absolute value gives you the size of a number — always positive.
We write $|k|$ for the absolute value of $k$.
(i) Why we need it — area
Look at this shape. The triangle has a piece above the $x$-axis and a piece below.
$A_1 = \dfrac{1}{2}(4)(3) = 6$ sq units
Above the axis — straightforward.
Now the bottom triangle. Base $= 4$, height $= -3$ (it's below the axis).
$\dfrac{1}{2}(4)(-3) = -6$ — but you can't have negative area.
Take the absolute value to throw away the sign:
$A_2 = \left|\dfrac{1}{2}(4)(-3)\right| = |-6| = 6$ sq units ✓
Absolute value
1.$|k|$ is the size of $k$.
2.Answer is always positive.
(ii) Quick examples
What is $|5|$?
$|5| = 5$
What is $|-5|$?
$|-5| = 5$
Both $5$ — it's the distance from $0$.
YOU TRY · 1
Find $|8|$.
Size of $8$ — it's already positive.
$|8| = 8$
$8$
YOU TRY · 2
Find $|-12|$.
Throw away the minus sign.
$|-12| = 12$
$12$
YOU TRY · 3
If $|k| = 7$, find $k$.
Two numbers have size $7$.
$|7| = 7$ and $|-7| = 7$, so $k = \pm 7$.
$k = \pm 7$
Section 2 of 5
Graphing y = |ax + b|
An absolute value graph is a V-shape sitting on the $x$-axis.
(i) Draw y = |x − 1|
Ask: where does the inside hit $0$?
$x - 1 = 0$
$x = 1$
So the V has its vertex at $x = 1$.
On the right of $x=1$, the graph is $y = x - 1$ (slope $m = 1$). On the left, reflect it up.
(ii) Draw y = |x + 2|
Where does the inside hit $0$?
$x + 2 = 0$
$x = -2$
Vertex at $x = -2$. Check two points either side:
$x = -3$: $y = |-3 + 2| = |-1| = 1$
$x = -1$: $y = |-1 + 2| = |1| = 1$
(iii) Draw y = |3x − 1|
Where does the inside hit $0$?
$3x - 1 = 0$
$3x = 1$
$x = \dfrac{1}{3}$
Vertex at $x = \dfrac{1}{3}$. The arms are steeper because slope $= 3$.
How to draw y = |ax + b|
1.Set $ax + b = 0$ to find the vertex $x$.
2.The V sits with its point on the $x$-axis there.
3.Plot two test points either side to set the steepness.
YOU TRY · 1
Where is the vertex of $y = |x - 4|$?
Set the inside to $0$.
$x - 4 = 0 \Rightarrow x = 4$
$x = 4$
YOU TRY · 2
Where is the vertex of $y = |x + 5|$?
Set the inside to $0$.
$x + 5 = 0 \Rightarrow x = -5$
$x = -5$
YOU TRY · 3
Where is the vertex of $y = |2x + 6|$?
Set the inside to $0$ then solve for $x$.
$2x + 6 = 0 \Rightarrow 2x = -6 \Rightarrow x = -3$
$x = -3$
Section 3 of 5
Solve graphically — plot and read
To solve $|ax+b| = c$, plot $y = |ax+b|$ and $y = c$. Read off the $x$-values where they meet.
(i) Solve |x + 2| = 1
Plot $y = |x + 2|$ (vertex at $x = -2$) and the horizontal line $y = 1$.
$x = -3$: $y = |-3 + 2| = 1$
$x = -1$: $y = |-1 + 2| = 1$
$x = -3$ OR $x = -1$
(ii) Solve |2x − 1| = 3
Plot $y = |2x - 1|$ (vertex at $x = \dfrac{1}{2}$) and the line $y = 3$.
$x = -1$ OR $x = 2$
Graphical method
1.Plot the V: $y = |ax + b|$.
2.Plot the horizontal line: $y = c$.
3.Read $x$ where they meet.
YOU TRY · 1
Solve $|x - 1| = 1$ graphically.
Vertex of V at $x = 1$. Check points where $|x - 1| = 1$.
$x = 0$: $|0-1| = 1$ ✓ $x = 2$: $|2-1| = 1$ ✓
$x = 0$ OR $x = 2$
YOU TRY · 2
Solve $|x + 3| = 2$ graphically.
Vertex at $x = -3$. Find where the V hits $y = 2$.
$x = -5$: $|-5+3| = 2$ ✓ $x = -1$: $|-1+3| = 2$ ✓
$x = -5$ OR $x = -1$
Section 4 of 5
Three methods — solve |x − 3| = 2
Here are three ways to solve $|x - 3| = 2$. Pick the one you trust.
(i) Method 1 — split with ±
$|k| = 2$ means $k = \pm 2$. So:
$x - 3 = 2$ OR $x - 3 = -2$
$x = 5$ OR $x = 1$
(ii) Method 2 — square both sides
Squaring kills the absolute value.
$(x - 3)^2 = 4$
$x^2 - 6x + 9 = 4$
$x^2 - 6x + 5 = 0$
$(x - 1)(x - 5) = 0$
$x = 1$ OR $x = 5$
(iii) Method 3 — diagram
Plot $y = |x - 3|$ and $y = 2$.
Vertex of V at $x = 3$. Plot points: $(2, 1)$, $(3, 0)$, $(4, 1)$.
$x = 1$ OR $x = 5$
Three methods for |ax + b| = c
1.Split: $ax + b = \pm c$
2.Square: $(ax + b)^2 = c^2$, then solve the quadratic.
3.Diagram: plot the V and the line $y = c$, read $x$.
YOU TRY · 1
Solve $|x - 1| = 2$ using $\pm$.
Split into two equations.
$x - 1 = 2$ OR $x - 1 = -2$ ⇒ $x = 3$ OR $x = -1$
$x = 3$ OR $x = -1$
YOU TRY · 2
Solve $|x + 4| = 3$ by squaring.
Square both sides, then expand and solve the quadratic.
$(x+4)^2 = 9$ ⇒ $x^2 + 8x + 16 = 9$ ⇒ $x^2 + 8x + 7 = 0$ ⇒ $(x+1)(x+7) = 0$
$x = -1$ OR $x = -7$
YOU TRY · 3
Solve $|2x + 1| = 5$ (any method).
Easiest: split with $\pm$.
$2x + 1 = 5$ ⇒ $x = 2$ OR $2x + 1 = -5$ ⇒ $x = -3$
$x = 2$ OR $x = -3$
Section 5 of 5
Two absolute values — |2x − 1| = |x + 2|
When BOTH sides have bars, plot both V's on the same diagram.
(i) Graphical method
$y = |2x - 1|$: vertex at $x = \dfrac{1}{2}$. Points $(0, 1)$ and $(1, 1)$.
$y = |x + 2|$: vertex at $x = -2$. Points $(-3, 1)$ and $(-1, 1)$.
$x = -\dfrac{1}{3}$ OR $x = 3$
(ii) Algebraic method
Drop the bars in two ways: same signs, or opposite signs.
$2x - 1 = x + 2$
$x = 3$
OR:
$2x - 1 = -(x + 2)$
$2x - 1 = -x - 2$
$3x = -1$
$x = -\dfrac{1}{3}$
For |A| = |B|
1.Solve $A = B$ AND $A = -B$.
2.Both solutions count.
YOU TRY · 1
Solve $|x| = |x - 4|$.
Two cases: same signs, opposite signs.
$x = x - 4$ ⇒ no solution. $x = -(x - 4)$ ⇒ $2x = 4$ ⇒ $x = 2$.
$x = 2$
YOU TRY · 2
Solve $|2x + 1| = |x - 3|$.
Same signs gives one $x$; opposite signs gives another.
$2x + 1 = x - 3$ ⇒ $x = -4$. $2x + 1 = -(x - 3) = -x + 3$ ⇒ $3x = 2$ ⇒ $x = \dfrac{2}{3}$.
$x = -4$ OR $x = \dfrac{2}{3}$
SUM
The lot in one box
Absolute value toolkit
1.$|k|$ is the size of $k$ — always positive.
2.$|k| = c$ ⇒ $k = \pm c$.
3.Graph $y = |ax + b|$: V-shape, vertex where $ax + b = 0$.
4.Solve $|ax + b| = c$: split with $\pm$, square, or diagram.
5.Solve $|A| = |B|$: $A = B$ AND $A = -B$.
End of lesson
Absolute Value — HL · Mathslive.ie