AP Precalculus: Parent Functions

To find an inverse by algebra: (1) write y = f(x), (2) swap x and y, (3) solve for y, and (4) name the result f−1(x). To verify, show that f(f−1(x)) = x and f−1(f(x)) = x. If a function is not one-to-one, restrict its domain first and keep the sign that matches the restriction. For a rational function, after swapping you need to clear the fraction, collect all terms containing y on one side, and factor y out.

1. Parent Functions, Piecewise Functions, and Function Notation

Every function in this unit is built from a small family of parent functions. Learn their shapes, domains, and ranges and the rest is shifting and stretching.

  • Constant y = c: domain all reals, range {c}
  • Linear y = x: domain and range all reals
  • Quadratic y = x²: domain all reals, range y ≥ 0
  • Cubic y = x³: domain and range all reals
  • Absolute value y = |x|: domain all reals, range y ≥ 0
  • Square root y = √x: domain x ≥ 0, range y ≥ 0
  • Reciprocal y = 1/x: domain and range all reals except 0
A piecewise function uses different rules on different parts of the domain. To evaluate it, first decide which interval the input belongs to, then use only that rule. Function notation works the same way: f(a) means replace every x with a, and f(x + h) means replace every x with the whole expression x + h. The difference quotient [f(x + h) − f(x)]/h measures an average rate of change and is simplified by expanding, canceling, and then dividing by h.

Worked Example: Let f(x) = x + 2 when x < 1 and f(x) = (x − 1)² when x ≥ 1. Find f(−2), f(1), and f(3).

−2 is less than 1, so use x + 2: f(−2) = 0.

1 satisfies x ≥ 1, so use (x − 1)²: f(1) = 0.

3 satisfies x ≥ 1: f(3) = 2² = 4.

f(−2) = 0, f(1) = 0, f(3) = 4

-5-4-3-2-112345-11234567xy
f(x) = x + 2 for x < 1 and (x − 1)² for x ≥ 1. The open circle at (1, 3) and the green point at (1, 0) show a jump discontinuity.

1. If f(x) = 2x² − 3x + 1, what is f(−2)?

Explanation: Replace x with −2 and use parentheses: 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 15. The common slip is forgetting that (−2)² is positive and −3(−2) is +6.

2. For f(x) = x² + 3x, what is the simplified difference quotient [f(x + h) − f(x)]/h?

Explanation: f(x + h) = (x + h)² + 3(x + h) = x² + 2xh + h² + 3x + 3h. Subtract f(x) = x² + 3x to get 2xh + h² + 3h. Dividing by h gives 2x + h + 3.

3. Let g(x) = 3x − 1 when x ≤ 2, and g(x) = x² − 5 when x > 2. What is g(2) + g(3)?

Explanation: Since 2 ≤ 2, g(2) = 3(2) − 1 = 5. Since 3 > 2, g(3) = 9 − 5 = 4. The sum is 9.

4. Which parent function has a domain of x ≥ 0 and a range of y ≥ 0?

Explanation: Only the square root function is limited to x ≥ 0 as inputs. Both y = x² and y = |x| accept every real input, and y = x³ has all real outputs.

2. Vertical and Horizontal Translations

A translation slides a graph without changing its shape. Adding a constant outside the function, y = f(x) + d, moves the graph up d units (down if d is negative). Adding a constant inside, y = f(x + c), moves the graph left c units, and f(x − c) moves it right c units. The horizontal direction feels backward, so think of it this way: the new graph needs input x = −c to do what the old graph did at 0. Every point (a, b) on y = f(x) lands at (a − c, b + d) on y = f(x + c) + d.

Worked Example: Describe how y = |x + 2| − 3 comes from y = |x|, and find the vertex.

The +2 is inside the absolute value, so the graph moves 2 units left. The −3 is outside, so it moves 3 units down.

The vertex of y = |x| is (0, 0), so the new vertex is (−2, −3).

Left 2, down 3; vertex (−2, −3)

-8-7-6-5-4-3-2-11234-5-4-3-2-112345xy
y = |x + 2| − 3 is y = |x| moved 2 units left and 3 units down, so the vertex moves from (0, 0) to (−2, −3).

1. What is the vertex of y = (x − 4)² + 1?

Explanation: x − 4 inside moves the parabola right 4, and +1 outside moves it up 1, so the vertex moves from (0, 0) to (4, 1).

2. Which equation moves y = √x left 5 units and down 2 units?

Explanation: Left 5 means x + 5 inside the radical, and down 2 means −2 outside. Choice D puts the 5 outside the radical, which would move the graph up.

3. The point (3, −2) is on the graph of y = f(x). Which point must be on y = f(x + 1) + 4?

Explanation: Replacing x with x + 1 moves points left 1, so x becomes 3 − 1 = 2. Adding 4 outside moves them up 4, so y becomes −2 + 4 = 2. The new point is (2, 2).

3. Dilations and Reflections

A vertical dilation y = a·f(x) multiplies every output by a: it stretches away from the x-axis when |a| > 1 and shrinks toward it when |a| < 1. If a is negative, the graph is also reflected over the x-axis. A horizontal dilation y = f(bx) divides every input by b: it compresses by a factor of 1/|b| when |b| > 1 and stretches by a factor of 1/|b| when |b| < 1. If b is negative, the graph is reflected over the y-axis. A point (a, c) on y = f(x) lands at (a/b, k·c) on y = k·f(bx).

Worked Example: Describe how y = (3x)² relates to y = x².

The 3 multiplies the input inside the function, so it is a horizontal change. A multiplier greater than 1 on the input compresses the graph by a factor of 1/3.

Check with a point: y = x² reaches height 9 at x = 3, while y = (3x)² reaches height 9 at x = 1.

Horizontal compression by a factor of 1/3

-3-2-1123-1123456789xyy = x²y = (3x)²
y = (3x)² is y = x² compressed horizontally by a factor of 1/3, so it reaches every height three times as fast.

1. The point (6, −4) is on y = f(x). Which point is on y = f(x/2)?

Explanation: f(x/2) is f(bx) with b = 1/2, a horizontal stretch by a factor of 2. Every x-coordinate doubles: 6 becomes 12. The y-coordinate does not change.

2. The point (2, 5) is on y = f(x). Which point must be on y = f(−x)?

Explanation: Replacing x with −x reflects the graph over the y-axis, so x changes sign and y stays the same. The point is (−2, 5). Reflecting over the x-axis would give (2, −5) instead, which comes from y = −f(x).

3. Which describes y = −2|x| compared with y = |x|?

Explanation: The −2 multiplies the output, so the change is vertical. The 2 stretches by a factor of 2 and the negative sign reflects over the x-axis, which turns the V upside down.

4. Combining Transformations, Domain, and Range

Write a transformed function as y = a·f(b(x + c)) + d. To build the graph, work in the same order you would evaluate: start inside the parentheses (horizontal shift and dilation), then apply the multiplier a (vertical dilation and reflection), and finish with the +d (vertical shift). To find the domain and range of a transformed graph, take the parent’s domain and range and apply the same moves: horizontal moves change the domain, and vertical moves change the range. A reflection or negative multiplier can flip the range, for example from y ≥ 0 to y ≤ 0.

Worked Example: Describe y = −3√(x + 1) as a transformation of y = √x, and state its domain and range.

Start inside the radical: x + 1 moves the graph 1 unit left. Next, the multiplier 3 is a vertical stretch by a factor of 3. Finally, the negative sign reflects over the x-axis.

The parent has domain x ≥ 0 and range y ≥ 0. Moving left 1 gives x ≥ −1. Stretching keeps the range at y ≥ 0, and reflecting flips it to y ≤ 0.

Domain x ≥ −1; range y ≤ 0

-3-2-11234567-9-8-7-6-5-4-3-2-112xy
y = −3√(x + 1) starts at (−1, 0) and passes through (0, −3) and (3, −6). Domain: x ≥ −1. Range: y ≤ 0.

1. What are the domain and range of y = 2√(x − 3) + 1?

Explanation: The radicand x − 3 must be at least 0, so x ≥ 3. The radical is at least 0, so 2√(x − 3) is at least 0, and adding 1 makes y ≥ 1.

2. What is the range of y = −2|x − 1| + 4?

Explanation: |x − 1| is at least 0, so −2|x − 1| is at most 0. Adding 4 gives y ≤ 4, with the maximum 4 at the vertex (1, 4).

3. Which sequence transforms y = √x into y = 3√(x + 4) − 2?

Explanation: Inside the radical, x + 4 moves the graph left 4. Then the multiplier 3 stretches it vertically, and −2 moves it down 2. Choice D reverses the order, and the stretch would then also scale the shift.

5. Function Models: Linear and Quadratic

Choosing a model means matching the story to a function type. A linear model fits a constant rate of change, such as a fixed fee plus a per-unit price. A quadratic model fits data that rise and fall symmetrically, or any area problem, and its vertex gives the maximum or minimum. Draw a diagram, define the variable, and restrict the domain to inputs that make sense in the situation. For a linear model from a table, check that the rate of change is constant, then write y − y1 = m(x − x1). For a quadratic y = ax² + bx + c, the vertex is at x = −b/(2a). With data and a calculator, a regression model finds the best-fit line or parabola.

Worked Example (linear): The price p (dollars) and quantity demanded q of an item are: (10, 90), (15, 80), (20, 70), (25, 60). Show q is linear in p, then write the model, interpret the slope, and find the implied domain.

The quantity drops by 10 every time the price rises by $5, so the slope is −10/5 = −2, a constant. Using (10, 90): q − 90 = −2(p − 10), so q = −2p + 110.

The slope means each $1 price increase lowers the quantity demanded by 2. The q-intercept is 110 and the p-intercept is 55 (set q = 0), so the implied domain is 0 ≤ p ≤ 55.

q = −2p + 110, with 0 ≤ p ≤ 55

Worked Example (quadratic): A ball is thrown so its height is h(t) = −5t² + 20t + 15 feet after t seconds. When is it highest, and how high?

The vertex has t = −20/(2(−5)) = 2. Then h(2) = −20 + 40 + 15 = 35.

Maximum height 35 feet at t = 2 seconds

1234551015202530354045xy
h(t) = −5t² + 20t + 15. The green point is the vertex (2, 35): the maximum height of 35 occurs at t = 2.

1. A plumber charges a $40 flat fee plus $25 per hour. What is the cost for 3.5 hours?

Explanation: The model is C(h) = 25h + 40. Then C(3.5) = 87.5 + 40 = 127.5. Choice B forgets the flat fee.

2. A ball’s height is h(t) = −5t² + 20t + 15. What is its maximum height?

Explanation: The vertex is at t = −20/(2(−5)) = 2 seconds, and h(2) = −20 + 40 + 15 = 35. The value 2 is when the maximum occurs, not how high it is.

3. A 24-inch by 18-inch sheet has a square of side x cut from each corner and the sides folded up to make an open box. If the base area must be 160 square inches, what is x?

Explanation: The base is (24 − 2x)(18 − 2x) = 160, which expands to 4x² − 84x + 432 = 160, or x² − 21x + 68 = 0. This factors as (x − 4)(x − 17) = 0. The side must satisfy 0 < x < 9, so x = 4. Check: 16 × 10 = 160.

4. Which situation is most likely to be modeled by a quadratic function?

Explanation: A fixed perimeter makes the area x(P/2 − x), which is quadratic with a maximum at its vertex. The other three have a constant rate of change, so they are linear.

6. Function Models: Piecewise, Polynomial, and Rational

A piecewise model fits a situation whose rule changes, such as tiered pricing or taxes. Identify the independent variable, write one rule for each interval, and be careful that the second rule only applies to the usage beyond the cutoff, which is x − cutoff. A polynomial model fits data with several zeros or turning points, and the zeros and their multiplicities suggest the factored form, such as d = k·t(t − a)² where a graph that only touches the axis at a signals an even multiplicity. A rational model fits inverse variation, y = k/x, where xy = k stays constant, so x1y1 = x2y2. Cost problems with a fixed area often produce rational functions such as 2x + A/x.

Worked Example: An electric company charges a $9 monthly fee plus $0.10 per kWh for the first 300 kWh and $0.07 per kWh after that. Write the monthly charge C(x) and find C(500).

For 0 ≤ x ≤ 300: C(x) = 9 + 0.10x. At x = 300 the charge is 9 + 30 = 39.

For x > 300 only the usage over 300 costs $0.07: C(x) = 39 + 0.07(x − 300).

For 500 kWh: C(500) = 39 + 0.07(200) = 39 + 14 = 53.

C(500) = $53

10020030040050060010203040506070xy
Monthly charge C(x). The slope drops from 0.10 to 0.07 at x = 300, so the graph bends at (300, 39).

1. If 12 workers can finish a job in 15 days, how many days will 9 workers need, assuming the same rate?

Explanation: Workers and days vary inversely, so 12 × 15 = 9 × d. Then d = 180/9 = 20 days. Fewer workers means more days, which rules out 11.25.

2. Parking costs $4 per hour for the first 2 hours and $2 per hour after that. What is the cost of 5 hours?

Explanation: The first 2 hours cost 2(4) = 8. The remaining 3 hours cost 3(2) = 6. The total is 14. Charging all 5 hours at $4 gives the incorrect $20.

3. A rectangular pen of area 200 square feet is built against a wall, so only three sides need fence. If x is the length of each side perpendicular to the wall, which function gives the total fence length?

Explanation: The side parallel to the wall has length 200/x because x × length = 200. The two perpendicular sides total 2x, and the wall side needs 200/x of fence, so F(x) = 2x + 200/x. Its domain is x > 0.

4. A graph touches the t-axis at t = 0, crosses it at t = 8 and stays above the axis between. Which form is most likely?

Explanation: Crossing the axis at both zeros means each has odd multiplicity, so the factors are t and (t − 8). That gives a quadratic that opens up or down. A squared factor would make the graph only touch the axis there.

7. Composition of Functions

The composite function (f ∘ g)(x) = f(g(x)) feeds the output of g into f. Work from the inside out: evaluate g first, then use that result as the input to f. To find the composite as an expression, replace every x in f with the entire expression for g. Composition is not commutative: f(g(x)) and g(f(x)) are usually different. The one special case is the identity function I(x) = x, which satisfies f(I(x)) = I(f(x)) = f(x). With tables, look up the inner output, then look that value up as the next input. With graphs, read the y-value of the inner function and use it as the x-value on the outer graph.

Worked Example: Let f(x) = 3x + 1 and g(x) = x² − 4. Find f(g(2)), f(g(x)), and g(f(x)).

g(2) = 4 − 4 = 0, then f(0) = 1.

f(g(x)) = f(x² − 4) = 3(x² − 4) + 1 = 3x² − 11.

g(f(x)) = g(3x + 1) = (3x + 1)² − 4 = 9x² + 6x − 3.

f(g(2)) = 1; f(g(x)) = 3x² − 11; g(f(x)) = 9x² + 6x − 3

-3-2-1123-12-8-44812162024xyf(g(x))g(f(x))
f(g(x)) = 3x² − 11 and g(f(x)) = 9x² + 6x − 3 are different functions, so order matters.

1. If f(x) = 2x − 5 and g(x) = x², what is f(g(3))?

Explanation: g(3) = 9 first, then f(9) = 2(9) − 5 = 13. Choice B is g(f(3)) = (1)² = 1, with the order reversed.

2. If f(x) = x + 4 and g(x) = 3x, which expression equals g(f(x))?

Explanation: g(f(x)) = g(x + 4) = 3(x + 4) = 3x + 12. Note that f(g(x)) = 3x + 4 is a different function.

3. If f(x) = x² and g(x) = x + 2, for what value of x does f(g(x)) = g(f(x))?

Explanation: f(g(x)) = (x + 2)² = x² + 4x + 4 and g(f(x)) = x² + 2. Setting them equal gives 4x + 4 = 2, so x = −1/2.

4. If f(x) = x² + 1, what is f(I(x)) where I(x) = x is the identity function?

Explanation: Composing with the identity function changes nothing: f(I(x)) = f(x) = x² + 1.

8. Domain of a Composite and Decomposing Functions

The domain of f(g(x)) contains the inputs x that are in the domain of g and whose output g(x) is in the domain of f. So find the domain of g, then remove any x for which g(x) is a value that f cannot accept. Decomposing a function runs composition backward: spot the “inside” expression (what sits in a radical, a denominator, an absolute value, or an exponent) and call it g(x), then call the remaining operation f. A function can be decomposed in several valid ways, but the simplest is usually best.

Worked Example: Let f(x) = 1/(x − 2) and g(x) = 3/(x + 1). Find the domain of f(g(x)).

The domain of g excludes x = −1. Next, f cannot accept 2, so find where g(x) = 2: 3/(x + 1) = 2 gives x + 1 = 3/2, so x = 1/2.

Check by simplifying: f(g(x)) = 1/(3/(x + 1) − 2) = (x + 1)/(1 − 2x), which is undefined at x = 1/2.

Domain: all real x except x = −1 and x = 1/2

Decomposing: For c(x) = √(5x − 3), the outer operation is the square root and the inside is 5x − 3, so g(x) = 5x − 3 and f(x) = √x.

1. If f(x) = √x and g(x) = x − 6, what is the domain of f(g(x))?

Explanation: f(g(x)) = √(x − 6), so x − 6 ≥ 0, which gives x ≥ 6.

2. If f(x) = 1/(x − 4) and g(x) = x + 3, which value is excluded from the domain of f(g(x))?

Explanation: f(g(x)) = 1/((x + 3) − 4) = 1/(x − 1), which is undefined at x = 1. The value x = 7 is where g(x) = 10, which is fine.

3. Which pair f and g gives f(g(x)) = |4x − 7| + 2?

Explanation: The inside expression is 4x − 7, so g(x) = 4x − 7. The outside operation is “absolute value, then add 2,” so f(x) = |x| + 2. Then f(g(x)) = |4x − 7| + 2.

9. Inverse Functions, One-to-One, and the Horizontal Line Test

The inverse function f−1 undoes f. If f(a) = b, then f−1(b) = a, so every ordered pair (a, b) of f becomes (b, a) for the inverse. The domain and range swap too. A function has an inverse that is also a function only if it is one-to-one: different inputs always give different outputs. Graphically, use the horizontal line test: if every horizontal line meets the graph at most once, f is one-to-one. The graph of f−1 is the reflection of the graph of f over the line y = x. If a function is not one-to-one, you can restrict its domain to a piece that is, such as x ≥ 0 for y = x².

Worked Example: The function f is {(−2, −7), (−1, 0), (0, 1), (1, 2), (2, 9)}. Find the inverse and state both domains.

Swap each pair: {(−7, −2), (0, −1), (1, 0), (2, 1), (9, 2)}.

Domain of f is {−2, −1, 0, 1, 2} and its range is {−7, 0, 1, 2, 9}. The inverse swaps them, so its domain is {−7, 0, 1, 2, 9} and its range is {−2, −1, 0, 1, 2}. All outputs are different, so f is one-to-one and the inverse is a function.

These pairs come from f(x) = x³ + 1, whose graph is shown with its inverse.

-4-3-2-11234-4-3-2-11234xy
f(x) = x³ + 1 and its inverse are mirror images over the dashed line y = x. The green points (0, 1) and (1, 0) are swapped copies of each other.

1. If f is invertible, f(5) = 9, and f(1) = 3, what is f−1(9)?

Explanation: f(5) = 9 means the inverse sends 9 back to 5, so f−1(9) = 5. The information f(1) = 3 is not needed.

2. Which set of ordered pairs describes a one-to-one function?

Explanation: A one-to-one function never repeats an output. Choice A has outputs 5, 6, 7, 8, which are all different. B repeats 4 and 1, C repeats 1, and D repeats 2.

3. Which function passes the horizontal line test on all real numbers?

Explanation: A line like y = 4 meets the graphs of x², |x|, and x4 at two points, since f(−a) = f(a) for each. The cubic is always increasing, so every horizontal line meets it exactly once.

4. The point (3, −2) is on the graph of f. Which point is on the graph of f−1?

Explanation: Inverse functions swap the coordinates, so (3, −2) becomes (−2, 3). This is a reflection over the line y = x.

10. Finding Inverses Algebraically

To find an inverse by algebra: (1) write y = f(x), (2) swap x and y, (3) solve for y, and (4) name the result f−1(x). To verify, show that f(f−1(x)) = x and f−1(f(x)) = x. If a function is not one-to-one, restrict its domain first and keep the sign that matches the restriction. For a rational function, after swapping you need to clear the fraction, collect all terms containing y on one side, and factor y out.

Worked Example: Find the inverse of f(x) = (2x + 5)/(x − 1).

Swap: x = (2y + 5)/(y − 1). Multiply: xy − x = 2y + 5. Collect y terms: xy − 2y = x + 5. Factor: y(x − 2) = x + 5, so y = (x + 5)/(x − 2).

Check: f(f−1(x)) = [2(x + 5)/(x − 2) + 5] / [(x + 5)/(x − 2) − 1] = (7x/(x − 2)) / (7/(x − 2)) = x.

f−1(x) = (x + 5)/(x − 2)

-8-6-4-22468-8-6-4-22468xy
f(x) = (2x + 5)/(x − 1) and f−1(x) = (x + 5)/(x − 2). Both are mirror images over y = x, and the asymptotes x = 1 and y = 2 of f become y = 1 and x = 2 for the inverse.

1. What is the inverse of f(x) = 4x − 9?

Explanation: Swap to get x = 4y − 9. Add 9: x + 9 = 4y. Divide by 4: y = (x + 9)/4.

2. The function f(x) = x² + 3 is restricted to x ≥ 0. What is f−1(x), and what is its domain?

Explanation: Swap: x = y² + 3, so y² = x − 3 and y = ±√(x − 3). The original domain x ≥ 0 becomes the inverse’s range, so choose the positive root. The inverse’s domain is the range of f, which is y ≥ 3.

3. What is the inverse of f(x) = 3/(x + 2)?

Explanation: Swap: x = 3/(y + 2). Then y + 2 = 3/x, so y = 3/x − 2 = (3 − 2x)/x.

4. Let f(x) = 2x + 6 and g(x) = (x − 6)/2. Which statement is true?

Explanation: f(g(x)) = 2((x − 6)/2) + 6 = x − 6 + 6 = x, and g(f(x)) = (2x + 6 − 6)/2 = x. Both compositions equal x, so they are inverses.

Common Mistakes to Avoid

  • Shifting the wrong way horizontally. f(x + 3) moves the graph left 3, and f(x − 3) moves it right 3.
  • Mixing up inside and outside changes. Constants and multipliers inside the parentheses change x-values (horizontal), and those outside change y-values (vertical).
  • Forgetting that a negative multiplier reflects. −f(x) flips over the x-axis, while f(−x) flips over the y-axis.
  • Evaluating a composition in the wrong order. f(g(x)) means do g first. Reversing it usually gives a different answer.
  • Dropping parentheses when substituting. For f(x) = x² − 3x, f(−2) = (−2)² − 3(−2), not −2² − 3(2).
  • Using the second rule for the whole input of a piecewise model. After a cutoff, only the amount beyond the cutoff gets the new rate: x − cutoff.
  • Finding an inverse of a function that is not one-to-one. Check the horizontal line test, or restrict the domain, before swapping x and y.

Frequently Asked Questions

How do I remember which direction a horizontal shift goes?

Ask what input makes the inside equal 0. In f(x + 3), the inside is 0 at x = −3, so the point that used to be at 0 is now at −3, which is 3 units to the left.

Is f −1(x) the same as 1/f(x)?

No. The −1 in f−1 means the inverse function, which undoes f. The reciprocal 1/f(x) is written with a fraction or as (f(x))−1.

How can I tell if two functions are inverses?

Compose them in both orders. If f(g(x)) = x and g(f(x)) = x for all inputs, they are inverses. Graphically, their graphs are reflections of each other over the line y = x.

Where can I practice more problems like these?

The Parent Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.