Look for this. If one machine uses another machine’s output, follow the input through both. If two quantities use the same input, evaluate them separately before combining them.
Write the inner rule in the input slot of the outer rule. Every allowed x must pass both stages. For a quotient, also remove every x that makes the divisor zero.
Worked example
Sketch two line segments on one grid. Graph f joins (0, 1) to (4, 5). Graph g joins (0, 4) to (4, 0). Read (f + g)(1) and f(g(1)). Which inputs can be used for f ∘ g?
At x = 1 the heights are f(1) = 2 and g(1) = 3. Add the heights to get 5. For composition, carry the height 3 across to the input axis. The f graph at 3 has height 4. Every g output is between 0 and 4, where f has a point. The composition domain is [0, 4], or {x ∈ ℝ | 0 ≤ x ≤ 4} in set-builder notation.
Try it yourself
Sketch a new pair. Graph p is the segment joining (0, 1) to (3, 4). Graph q joins (0, −2) to (4, 2). Read p(q(3)), and find the domains of p ∘ q and the pointwise quotient p/q.
Check your reasoning
At x = 3, q has height 1. The p graph at input 1 has height 2, so p(q(3)) = 2. For p ∘ q, q must supply an input in [0, 3]. Its height first reaches 0 at x = 2 and reaches 2 at its last point, so the domain is [2, 4]. For p/q, both graphs have heights on [0, 3]. Exclude x = 2 where q is zero, leaving {x ∈ ℝ | 0 ≤ x ≤ 3, x ≠ 2}.
Outcome RF2
Follow the endpoint through a translation
Look for this. An added constant inside or outside a rule can move a whole graph without changing its shape.
In y = f(x − h) + k, a point (u, v) moves to (u + h, v + k). Solve x − h = u to see why a positive h moves the point right. Apply the same move to a domain boundary.
Worked example
Translate y = √x so that its endpoint moves from (0, 0) to (3, −2). Write the equation, locate its x-intercept and describe its range.
The new rule is y = √(x − 3) − 2. At y = 0 the square root equals 2, so x − 3 = 4 and the intercept is (7, 0). The endpoint is its lowest point, so y ≥ −2.
Try it yourself
A temperature trace f has domain [0, 8] hours and a peak at (5, 19). A second recording shows the same trace 2 hours earlier and 4 degrees warmer. Write its rule, domain and peak.
First-move hint
The peak used to happen at hour 5. Where would it move on a timeline if it happened two hours earlier?
Check your reasoning
At time x, use the old reading at x + 2 and add 4. The new rule is f(x + 2) + 4. Both time limits shift left, giving [−2, 6]. The peak moves to (3, 23).
Outcome RF3
Measure a stretch from the axes
Look for this. A multiplier inside the input changes horizontal distances. A multiplier outside changes heights.
For y = a f(bx), solve bx = u for the new x-coordinate. This sends (u, v) to (u/b, av). The horizontal scale factor is 1/|b| and the vertical scale factor is |a|.
Worked example
A recorded pulse has a peak at (6, 8) and domain [0, 12]. Find the peak and domain after changing the rule to y = ½ f(3x).
The original peak appears when 3x = 6. Halving its height gives the point (2, 4). Divide both domain endpoints by 3 to get [0, 4].
Try it yourself
A graphing program should move a marked point (4, 3) to (8, 12) using only a horizontal stretch and a vertical stretch. Enter the transformed rule in terms of f.
Check your reasoning
Horizontal distances double because 8/4 = 2, so the input must be x/2. Heights scale by 12/3 = 4. The rule y = 4f(x/2) gives 4f(4) = 12 at x = 8.
Outcome RF4
Build a graph from a point map
Look for this. When a rule contains both stretches and translations, use one coordinate map for every point.
For y = a f(b(x − h)) + k, solve b(x − h) = u to get x = u/b + h, then use y = av + k. Factor the input first if the horizontal shift is hard to see.
Worked example
A piecewise graph consists of line segments joining (−2, 0), (0, 4) and (4, 2). Sketch y = 3f(2x − 6) − 1. State its domain and range.
The input is 2(x − 3), so map (u, v) to (u/2 + 3, 3v − 1). Join (2, −1), (3, 11) and (5, 5) in that order with straight segments. The domain is [2, 5] and the range is [−1, 11], or {y ∈ ℝ | −1 ≤ y ≤ 11}.
Try it yourself
A parabola starts as f(x) = x². Its horizontal and vertical distances from the axes are first multiplied by 3 and 2 respectively. Then it moves 1 left and 4 up. Find its new rule and the images of its vertex and (3, 9).
Check your reasoning
The coordinate map is (u, v) → (3u − 1, 2v + 4). Reversing the horizontal map gives y = 2f((x + 1)/3) + 4. The vertex moves to (−1, 4) and (3, 9) moves to (8, 22). Substituting x = 8 gives y = 22.
Outcome RF5
Identify the mirror line from what stays fixed
Look for this. A mirror keeps points on its line fixed. Across y = x, input and output trade places.
Across the x-axis use (u, −v), across the y-axis use (−u, v), and across y = x use (v, u). Translate the point map into an equation by changing the corresponding coordinate.
Worked example
A trail profile follows y = √x and passes through (9, 3). Describe separately its reflection in each coordinate axis, including the new point and equation.
Across the x-axis the height changes sign. The rule is y = −√x and the point is (9, −3). Across the y-axis the input changes sign. The rule is y = √(−x) with x ≤ 0 and the point is (−9, 3). Both graphs keep the origin.
Try it yourself
Reflect the segment from (−1, 2) to (3, 6) in the line y = x. Give the endpoints, the equation on its new domain, and whether any point stayed fixed.
Check your reasoning
Swap each ordered pair to get (2, −1) and (6, 3). The original segment satisfies y = x + 3. Swapping x and y gives x = y + 3, so the new segment is y = x − 3 on [2, 6]. No original point has equal coordinates, so none stayed fixed.
Outcome RF6
Choose a branch before undoing a quadratic
Look for this. When two inputs give the same output, the inverse relation needs a branch choice to be a function.
Swap x and y, solve for y, then use the original input restriction to choose the sign. The original range becomes the inverse domain. Check by tracing a point there and back.
Worked example
An instrument uses f(t) = (t − 3)² + 2 for t ≥ 3. A reading is 18. Derive the inverse and recover t.
From x = (y − 3)² + 2 we get y = 3 ± √(x − 2). Since t ≥ 3, take the plus branch. Thus f⁻¹(x) = 3 + √(x − 2) for x ≥ 2. The reading gives t = 3 + √16 = 7.
Try it yourself
For g(x) = 9 − x², a student writes g⁻¹(x) = √(9 − x). Explain the needed domain restriction and give the inverse instead for a device that uses x ≤ 0.
Check your reasoning
Across all real inputs, g(−2) and g(2) both equal 5, so that inverse relation has two branches. The student's positive square root requires original x ≥ 0. For the device choose g⁻¹(x) = −√(9 − x), with domain x ≤ 9 and range y ≤ 0.
Worked notes and diagrams
1
The words, first
The idea: describe a transformation by naming its direction, scale factor or distance, and how it moves a point.
The dashed curve is the parent and the solid one its image. h and k are the journey of a single point; a and b change the shape around it.
Word
What it means
Parent function
The plain version of a graph before anything is done to it: y = x², y = √x, y = |x|.
Image
Where a point ends up after the transformation; the original is the pre-image.
Mapping notation
A rule of the form (x, y) → (new x, new y).
Translation
A slide, horizontally by h or vertically by k. The shape does not change.
Stretch
A scaling away from an axis — vertical by |a|, horizontal by 1/|b|.
Compression
The same operation with a factor below 1.
Reflection
A flip across an axis: a < 0 across the x-axis, b < 0 across the y-axis.
Invariant point
A point the transformation leaves exactly where it was.
Inverse
The inverse relation swaps each (x, y) to (y, x). It is a function, written f⁻¹, when each original output has one input.
2
Translations and stretches
The idea: in y = a·f(b(x − h)) + k, anything outside the function changes y and does what it looks like. Anything inside, with x, changes x and does the opposite of what it looks like.
x − 3 looks like "minus" but moves the graph right. The + 1 outside moves it up, exactly as it looks.
In the equation
What happens
y = f(x) + k
up k (down if k is negative)
y = f(x − h)
right h (left if written x + h)
y = a·f(x)
vertical stretch by a factor of |a|
y = f(bx)
horizontal stretch by a factor of 1/|b|
The mapping rule
For y = a·f(b(x − h)) + k, every point moves like this:
(x, y) → (x/b + h, a·y + k)
Stretch first, then translate. For y = 3f(x − 1) + 2, the point (4, −6) becomes (4 + 1, 3 × (−6) + 2) = (5, −16).
Watch out: y = f(2x − 6) is not a shift of 6. Factor out the 2 first: f(2(x − 3)), so the shift is 3 right.
3
Reflections and inverses
The idea: a negative outside flips y; a negative inside flips x; swapping x and y reflects the whole graph in the line y = x — that is the inverse.
Equation
Reflection in
(3, 4) becomes
y = −f(x)
the x-axis
(3, −4)
y = f(−x)
the y-axis
(−3, 4)
x = f(y), the inverse
the line y = x
(4, 3)
Finding an inverse algebraically
For f(x) = 2x − 6: write y = 2x − 6, swap to x = 2y − 6, and solve for y: f⁻¹(x) = (x + 6)/2. Notice it undoes the original steps in reverse — add 6, then divide by 2.
When the inverse is not a function
Reflect y = x² in y = x and you get a sideways parabola, which fails the vertical line test. Restricting the original domain to x ≥ 0 keeps only half, and then the inverse, y = √x, is a function.
Watch out: f⁻¹(x) does not mean 1/f(x). The −1 is a label for "inverse", not an exponent.
4
Combining functions
The idea: (f + g)(x), (f − g)(x), (f·g)(x) and (f/g)(x) combine the two heights at the same x-coordinate. Check that both heights exist and, for division, that the second height is nonzero.
With f(x) = x² − 4 and g(x) = x − 2:
(f + g)(x) = x² + x − 6
(f − g)(x) = x² − x − 2
(f·g)(x) = (x² − 4)(x − 2) = x³ − 2x² − 4x + 8
(f/g)(x) = (x − 2)(x + 2)/(x − 2) = x + 2, x ≠ 2
For pointwise addition, subtraction and multiplication, the domain contains the inputs where both original functions are defined. For a quotient, you also remove every x that makes g(x) = 0 — and that restriction stays even after the factor cancels.
Watch out: simplifying f/g to x + 2 and forgetting x ≠ 2. The graph has a hole there.
5
Composition
The idea: f(g(x)) means put x into g, then put that answer into f. Always work from the inside out.
Worked example
f(x) = x + 1 and g(x) = 2x. Find f(g(3)) and g(f(3)).
f(g(3)): g(3) = 6, then f(6) = 7.
g(f(3)): f(3) = 4, then g(4) = 8.
Different orders, different answers. Composition is usually not the same both ways.
As expressions
With f(x) = x² and g(x) = x + 3: f(g(x)) = (x + 3)² but g(f(x)) = x² + 3.
Domain of a composition
f(g(x)) is only defined where g(x) is defined and its output is allowed into f. With f(x) = √x and g(x) = x − 5, f(g(x)) = √(x − 5), so x ≥ 5.
Working backwards
To write h(x) = (2x − 1)² as a composition, find the steps: first 2x − 1, then square. The first step is the inside function: g(x) = 2x − 1 and f(x) = x².
Watch out: reading f(g(x)) left to right and doing f first. The function closest to x always goes first.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Translations and stretches · 6 questions
Practice 1
The graph of y = f(x) is translated 3 units to the right. What is the new equation?
y = f(x) − 3
y = f(x − 3)
y = f(x) + 3
y = f(x + 3)
Answer and explanation
y = f(x − 3)
Horizontal shifts act on x and go the opposite way to the sign: x − 3 moves the graph right. Adding 3 outside the function would move it up.
Practice 2
Compared with y = f(x), the graph of y = f(x) + 5 is:
translated 5 units down
translated 5 units left
translated 5 units right
translated 5 units up
Answer and explanation
translated 5 units up
Adding outside the function adds to every y-value, so the whole graph moves up 5. Vertical shifts go the same way as the sign.
Practice 3
Compared with y = f(x), the graph of y = f(2x) has:
a vertical stretch by a factor of 2
a horizontal stretch by a factor of 2
a horizontal stretch by a factor of 1/2
a vertical stretch by a factor of 1/2
Answer and explanation
a horizontal stretch by a factor of 1/2
A number multiplying x stretches horizontally by its reciprocal, so 2x squeezes the graph toward the y-axis by a factor of 1/2.
Practice 4
The point (4, −6) is on y = f(x). Which point is on y = 3f(x − 1) + 2?
(13, −16)
(3, −16)
(5, −16)
(5, −12)
Answer and explanation
(5, −16)
x: shift right 1, so 4 + 1 = 5. y: stretch first, then shift — 3 × (−6) + 2 = −16. Shifting before stretching gives the wrong −12.
Practice 5
The point (6, 3) is on y = f(x). Which point is on y = −2f(½(x + 4))?
(8, 6)
(8, −6)
(−1, −6)
(16, −6)
Answer and explanation
(8, −6)
x: the ½ stretches horizontally by 2, then shift left 4 — 6 × 2 − 4 = 8. y: stretch by 2 and reflect — 3 × (−2) = −6. Stretches come before translations.
Practice 6
y = f(x) has domain −2 ≤ x ≤ 6. What is the domain of y = f(2x)?
0 ≤ x ≤ 8
−2 ≤ x ≤ 6
−1 ≤ x ≤ 3
−4 ≤ x ≤ 12
Answer and explanation
−1 ≤ x ≤ 3
f(2x) compresses horizontally by 1/2, so every x-value is halved: −2 becomes −1 and 6 becomes 3. Doubling them is the common reversal.
Reflections and inverses · 6 questions
Practice 1
The graph of y = −f(x) is a reflection of y = f(x) in:
the origin
the y-axis
the x-axis
the line y = x
Answer and explanation
the x-axis
The negative is outside, so every y-value changes sign and the graph flips over the x-axis. y = f(−x) would flip it over the y-axis.
Practice 2
The graph of the inverse of a function is a reflection of the original graph in:
the x-axis
the line y = −x
the y-axis
the line y = x
Answer and explanation
the line y = x
An inverse swaps every x and y, and swapping coordinates is exactly a reflection in the line y = x.
Practice 3
The point (2, 5) is on the graph of y = f(x). Which point is on the graph of its inverse?
(−5, −2)
(2, −5)
(5, 2)
(−2, 5)
Answer and explanation
(5, 2)
An inverse swaps x and y, so (2, 5) becomes (5, 2).
Practice 4
What is the inverse of f(x) = 2x − 6?
f⁻¹(x) = (x + 6)/2
f⁻¹(x) = (x − 6)/2
f⁻¹(x) = 2x + 6
f⁻¹(x) = 1/(2x − 6)
Answer and explanation
f⁻¹(x) = (x + 6)/2
Swap x and y: x = 2y − 6. Solve for y: y = (x + 6)/2. The inverse undoes the steps in reverse order — add 6, then divide by 2.
Practice 5
The point (3, 4) is on y = f(x). Which point is on y = f(−x)?
(−3, −4)
(4, 3)
(−3, 4)
(3, −4)
Answer and explanation
(−3, 4)
The negative is inside with x, so only x changes sign: a reflection in the y-axis. (3, −4) would be y = −f(x).
Practice 6
Which domain for f(x) = x² gives a one-to-one function while preserving its full range y ≥ 0?
no restriction is needed
y ≥ 0
x ≥ 1
x ≥ 0
Answer and explanation
x ≥ 0
On x ≥ 0, each non-negative output has precisely one input. The domain x ≥ 1 loses outputs below 1. The unrestricted parabola assigns the same output to opposite nonzero inputs, and y ≥ 0 describes outputs.
Combining functions · 2 questions
Practice 1
If f(x) = 2x and g(x) = x + 3, what is (f + g)(x)?
3x + 3
2x² + 6x
3x
2x + 3
Answer and explanation
3x + 3
(f + g)(x) just means f(x) + g(x): 2x + x + 3 = 3x + 3. Multiplying them would give 2x² + 6x.
Practice 2
If f(x) = x² − 4 and g(x) = x − 2, what is (f/g)(x)?
x + 2, with no restriction
x − 2, x ≠ 2
x + 2, x ≠ 2
x² − x − 2
Answer and explanation
x + 2, x ≠ 2
(x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2. But you can never divide by g(x) = 0, so x ≠ 2 stays even after cancelling.
Composition · 4 questions
Practice 1
If f(x) = x + 1 and g(x) = 2x, what is f(g(3))?
8
9
7
6
Answer and explanation
7
Work from the inside out: g(3) = 6, then f(6) = 7. Doing f first gives g(f(3)) = 8, which is a different composition.
Practice 2
If f(x) = x² and g(x) = x + 3, what is g(f(x))?
x³ + 3x²
x² + 3
(x + 3)²
x² + x + 3
Answer and explanation
x² + 3
f goes in first: g(x²) = x² + 3. (x + 3)² is the other order, f(g(x)) — composition is not usually the same both ways.
Practice 3
If f(x) = √x and g(x) = x − 5, what is the domain of f(g(x))?
all real numbers
x ≥ −5
x ≥ 0
x ≥ 5
Answer and explanation
x ≥ 5
f(g(x)) = √(x − 5), and the expression under the root cannot be negative: x − 5 ≥ 0, so x ≥ 5.
Practice 4
h(x) = (2x − 1)². Which pair of functions gives h(x) = f(g(x))?
f(x) = x², g(x) = 2x − 1
f(x) = 2x, g(x) = x² − 1
f(x) = x − 1, g(x) = 4x²
f(x) = 2x − 1, g(x) = x²
Answer and explanation
f(x) = x², g(x) = 2x − 1
The inside step is 2x − 1, so that is g. The outside step squares the result, so that is f. Swapping them — f(x) = 2x − 1 and g(x) = x² — builds 2x² − 1 instead.
maddyhelps.