6.4 Transformations and Interpreting Functions in Context
Key Takeaways
f(x)+k shifts the graph vertically (up if k>0); f(x−h) shifts it right by h, so f(x+3) is left 3.
a f(x) stretches vertically if |a|>1 and reflects across the x-axis if a<0; f(−x) reflects across the y-axis.
If g(x)=3(x+8), then g(4)=3(4+8)=36, not 3(4)+8=20.
In context, h(3)=18 names an output at input 3, while h(t)+2 is two units higher than h(t) and h(t+2) is two input units later.
Describe a slide or flip in function notation, such as 2f(x−3)+1, not only in words.
6.4 Transformations and Interpreting Functions in Context
The last Table 11 Functions skill is interpreting functions in a context — reading f(3) = 18 as “after 3 hours the tank holds 18 gallons,” not as a floating algebra trick. Higher Skills Insight bands reward fluency using function notation to describe transformations: writing f(x − 4) + 2 for a shift, rather than only sliding a sketch — the 276–300 absolute-value item is exactly that, in function notation. This section ties those together, with piecewise rules at the level AAF actually uses.
Functions is 10–20% of AAF (typically 2–4 CAT items). Practice the same notation on FREE items at /practice/accuplacer-advanced-algebra. Parabola slides and vertex form are the quadratic special case in Quadratic Forms, Graphs, Vertex, and Intercepts. Official skill names are on College Board’s What’s on the Tests page.
The five transformations you must name
Start with a parent graph y = f(x). New graphs are written by composing simple changes with f.
| Notation | Geometric effect | Example with f(x) = |x| |
|---|---|---|
f(x) + k | Vertical shift up k (down if k < 0) | |x| + 3 moves the V up 3 |
f(x − h) | Horizontal shift right h (left if h < 0) | |x − 4| moves the V right 4 |
a f(x) | Vertical stretch if |a| > 1; vertical shrink if 0 < |a| < 1; x-axis reflection if a < 0 | 3|x| is steeper; −|x| opens down |
f(−x) | Reflection across the y-axis | |−x| = |x| (this parent is already symmetric) |
−f(x) | Reflection across the x-axis | −|x| is an upside-down V |
Two facts cause most errors.
Horizontal substitutions move the opposite direction from the sign you see. f(x − 3) is a shift right 3, because x − 3 = 0 when x = 3: the old input 0 now happens at 3. f(x + 3) = f(x − (−3)) is a shift left 3. Vertical changes + k move the way the sign looks: +2 is up, −2 is down.
Inside versus outside. Anything outside f, such as 2f(x) or f(x)+5, changes outputs (y). Anything inside the argument, such as f(2x) or f(x−5), changes inputs (x). f(2x) is a horizontal shrink by 2, not a vertical stretch. AAF is more likely to test f(x−h) and af(x) than the f(2x) shrink, but the inside/outside split still saves you.
Combined: g(x) = 2f(x − 3) + 1 is right 3, vertical stretch 2, up 1, in that mental order: horizontal first (inside), then stretch, then vertical shift (outside). If f is |x|, the vertex moves from (0, 0) to (3, 1) and the arms get steeper. If f is x^2, the vertex of 2(x − 3)^2 + 1 is likewise (3, 1) — the same transformation language as vertex form.
f(−x) versus −f(x): for f(x) = x^3 + 1, f(−x) = −x^3 + 1 (y-axis reflection) and −f(x) = −x^3 − 1 (x-axis reflection). They are not the same function. Describing “flip over the x-axis” in notation is −f(x), not f(−x). That sentence is the higher-band transformation fluency item.
A vertical shrink example: (1/2)f(x) halves every output. If f(2) = 10, then (1/2)f(2) = 5. A horizontal counterpart f(x/2) would wait until the input is twice as large to produce the old output; do not mix those.
Worked substitution: g(x) = 3(x + 8)
Context items often hide a transformation inside a formula that looks too simple. Let g(x) = 3(x + 8). This is a linear function, equivalently g(x) = 3x + 24. To evaluate, substitute the entire input into (x + 8) first.
g(4) = 3(4 + 8) = 3(12) = 36.
Traps:
3(4) + 8 = 20treats the 8 as added after multiplying, as if the rule were3x + 8.3(4)alone is 12 — dropping the inner+8entirely.3(12) + 8 = 44substitutes correctly, then adds 8 again as ifgwere3(x+8)+8.- Distribution
3(4) + 3(8) = 12 + 24 = 36is valid, but only if you distribute to both terms inside.
If a story says a lab records temperature g(t) = 3(t + 8) degrees after t minutes past dawn, then at t = 4 the temperature is 36, not 20. The +8 is inside the input to the scale factor 3: eight minutes of warm-up are built into the clock. Interpreting in context means attaching units and the correct substitution, not inventing a new formula.
Another original context: a parking garage charges C(h) = 2(h − 1) + 5 dollars for h hours after the first hour is bundled into a $5 gate fee. C(4) = 2(3) + 5 = 11. Writing 2(4 − 1) + 5 is the function; writing 2(4) − 1 + 5 = 12 drops the parentheses and misreads the story. C(1) = 5, the gate fee with no extra hours billed.
Piecewise functions at AAF level
A piecewise function uses different formulas on different parts of the domain. AAF pieces are short: two linear pieces, or a linear piece glued to a constant, or the definition of absolute value.
Let p(x) = x + 2 when x < 0, and p(x) = 2x when x ≥ 0.
Then p(−3) = −3 + 2 = −1 (left piece) and p(4) = 8 (right piece). p(0) = 0 because 0 uses x ≥ 0. The graph is two rays that meet at the origin in this particular example; they need not meet. If the pieces disagree at a boundary, the inequality (< versus ≤) tells you which value is defined. AAF will not usually ask you to debate a single missing point, but it will ask you to pick the correct piece.
Absolute value is piecewise: |x| = −x for x < 0 and x for x ≥ 0. Transformations of |x| are still piecewise after you shift: |x − 4| + 1 uses 5 − x to the left of 4 and x − 3 to the right of 4. The V-graph in Graphs of Functions, Vertical Line Test, Maxima and Minima is this piecewise rule drawn.
For a shipping-style context: a mailer charges $4 for weights up to 2 lb and $4 + 1.5(w − 2) for w > 2. That is piecewise. S(2) = 4 and S(6) = 4 + 1.5(4) = 10. Do not apply the second piece at w = 2 if the story says “over 2 lb.” Do not apply 1.5w to the whole weight; the 1.5 multiplies only the extra pounds.
Interpreting f in a sentence
When h(t) is height in meters after t seconds:
| Symbol | Meaning |
|---|---|
h(3) = 18 | Height is 18 m at 3 s |
h(t) = 18 | Solve for the time when height is 18 m |
h(t+2) | Height 2 seconds later than time t |
h(t)+2 | Height 2 meters higher than at time t |
h(3) − h(1) | Change in height from 1 s to 3 s |
h^{-1}(18) | The time at which height is 18 m (if one-to-one on the interval) |
That table is the same f(x+2) versus f(x)+2 distinction from Section 6.1, now with units. If a water tank has volume V(t) = 40 − 3t gallons after t minutes of draining, V(5) = 25 gallons remain, V(t) = 10 solves 40 − 3t = 10 so t = 10 minutes, and the tank is empty when V(t) = 0, t = 40/3 minutes. Domain in context is not all reals: t ≥ 0 and V(t) ≥ 0, so 0 ≤ t ≤ 40/3. Context restricts the algebraic domain.
Average rate of change on [1, 5] is (V(5) − V(1))/(5 − 1) = (25 − 37)/4 = −3 gallons per minute, which matches the slope of this linear model. AAF may not use the phrase “average rate of change,” but it will ask how fast a quantity changes between two inputs, or what V(8) means in gallons, not as a bare 16.
If a quadratic model h(t) = −5t^2 + 20t + 2 gives height of a thrown object, the vertex time t = −b/(2a) = 2 seconds is when height is maximum: h(2) = −20 + 40 + 2 = 22 meters. Interpreting that 22 as “the object travels 22 meters horizontally” is the wrong quantity. Function notation names one output of one rule; read the units on h.
After you can shift, reflect, substitute g(4) = 3(4+8), and read h(3) = 18 in words, move to mixed CAT practice. The Functions slice is the heaviest Table 10 weight on AAF — typically up to four of the twenty items — and it reuses the vertex skill you already built for quadratics.
If g(x) = 3(x + 8), what is g(4)?
20
12
36
44
Compared with y = f(x), the graph of y = f(x − 3) + 2 is which transformation?
Left 3, up 2
Right 3, down 2
Left 3, down 2
Right 3, up 2
The graph of y = −f(x) is which change to y = f(x)?
A reflection of y = f(x) across the x-axis
A reflection of y = f(x) across the y-axis
A shift of y = f(x) down by 1
A horizontal stretch of y = f(x)
Sections you finish are checked off in the contents.