Absolute Value Transformations

Absolute Value Transformations — Moosa Academy

The absolute value parent $f(x) = |x|$ is a V with its vertex at the origin. One general form, $f(x) = a|x - h| + k$ , carries every translation, stretch and reflection this V can undergo — and reading off its vertex and shape takes only a glance at $a$ , $h$ and $k$ .

Concept The general form

f(x) = a|x - h| + k

Vertex — always at

(h,\ k)

|a| — vertical stretch (

|a| > 1

) or compression (

0 < |a| < 1

Sign of a — negative reflects the V in the x-axis, so it opens downward.

As with every parent function, the sign inside the bracket is reversed: $h > 0$ moves the vertex right, and $h < 0$ moves it left. Outside the bracket, $k$ behaves as expected: positive lifts, negative lowers.

Example Reading off the vertex

Find the vertex of $f(x) = |x - 4| + 2$ .

Matching against

a|x - h| + k

gives

h = 4

and

k = 2

The vertex of any member of this family is

(h,\ k)

⟹ vertex (4, 2)

Example Reading off the shape

Describe $f(x) = -3|x|$ compared with the parent $|x|$ .

Here

a = -3

, so

|a| = 3 > 1

: the V is stretched vertically, narrower than the parent.

Since

a < 0

, the graph is reflected in the x-axis and opens downward.

⟹ a narrow V, opening downward, vertex still at the origin

Concept Solving absolute value equations

Because $|x|$ measures distance from zero, the equation $|x| = c$ (for $c > 0$ ) always has two solutions: the two points $c$ units from zero.

|\text{expression}| = c

splits into: expression

= c

, or expression

= -c

Graphically, this is exactly where the horizontal line $y = c$ crosses both arms of the V.

Example Solving an equation

Solve $|x + 2| = 7$ .

Case 1:

x + 2 = 7 \Rightarrow x = 5

Case 2:

x + 2 = -7 \Rightarrow x = -9

Check: both 5 and −9 are exactly 7 units from −2.

⟹ x = 5 or x = −9

Example Building the equation from a description

Write the function whose V has vertex $(-1,\ -3)$ and is twice as narrow as $|x|$ .

A negative

h = -1

becomes

+1

inside the bracket:

(x + 1)

k = -3

, added outside to lower the vertex.

Twice as narrow means

a = 2

⟹ f(x) = 2|x + 1| − 3

Checking: $f(-1) = 2|0| - 3 = -3$ , matching the required vertex.

Summary

In f(x) = a|x − h| + k, the vertex is always (h, k).
|a| stretches (|a| > 1) or compresses (0 < |a| < 1) the V; a negative a reflects it downward.
|expression| = c splits into two linear equations, expression = c and expression = −c.
The same reversed-sign rule for h applies here as for every other parent function.