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ACT · Math · Functions

Nonlinear functions, features and transformations

Connect rules, graphs and tables for nonlinear functions and choose forms that reveal the requested feature.

What you will tackle

  • Create and interpret quadratic and exponential models
  • Connect polynomial, rational, radical and exponential rules, tables and graphs
  • Interpret intercepts, vertices, growth factors and domains
  • Transform inputs and outputs
  • Recover inputs from outputs and connect roots with graph intercepts

Features depend on the representation

For f(x) = (x − 3)² − 4, the vertex is (3,−4), the roots are 1 and 5, and the y-intercept is 5. Factored form (x − 1)(x − 5) displays the roots, while vertex form displays the minimum. A rational graph may exclude inputs where its denominator is zero; a square-root graph has a restricted real domain.

Transformations change pairings

Replacing x by x − h shifts a graph h units right; adding k shifts it up k. In P(t) = 300(1.08)^t, 300 is the initial amount and 1.08 represents 8% growth each time period. Equal increments give a linear pattern, whereas equal ratios suggest an exponential pattern. A fitted rule still needs contextual validation.

Three original function tables
ModelInput/output pairs
P(−2,−8), (−1,−1), (0,0), (1,1), (2,8)
R(1,4), (2,2), (4,1)
S(0,0), (1,1), (4,2), (9,3)
Graph P · sampled polynomial curveP: (-2, -8); (-1.8, -5.832000000000001); (-1.6, -4.096000000000001); (-1.4, -2.7439999999999993); (-1.2, -1.728); (-1, -1); (-0.8, -0.5120000000000001); (-0.6, -0.216); (-0.4, -0.06400000000000002); (-0.2, -0.008000000000000002); (0, 0); (0.2, 0.008000000000000002); (0.4, 0.06400000000000002); (0.6, 0.216); (0.8, 0.5120000000000001); (1, 1); (1.2, 1.728); (1.4, 2.7439999999999993); (1.6, 4.096000000000001); (1.8, 5.832000000000001); (2, 8). Horizontal axis: x; vertical axis: P(x). Invented teaching data.-8-2-4-1004182xP(x) P
Graph P · sampled polynomial curve · original teaching data; the table/description supplies exact coordinates.
Graph R · positive-input portionR: (1, 4); (1.25, 3.2); (1.5, 2.6666666666666665); (1.75, 2.2857142857142856); (2, 2); (2.25, 1.7777777777777777); (2.5, 1.6); (2.75, 1.4545454545454546); (3, 1.3333333333333333); (3.25, 1.2307692307692308); (3.5, 1.1428571428571428); (3.75, 1.0666666666666667); (4, 1); (4.25, 0.9411764705882353); (4.5, 0.8888888888888888); (4.75, 0.8421052631578947); (5, 0.8). Horizontal axis: x; vertical axis: R(x). Invented teaching data.0011.2522.533.7545xR(x) R
Graph R · positive-input portion · original teaching data; the table/description supplies exact coordinates.
Graph S · nonnegative-input portionS: (0, 0); (0.25, 0.5); (0.5, 0.7071067811865476); (0.75, 0.8660254037844386); (1, 1); (1.25, 1.118033988749895); (1.5, 1.224744871391589); (1.75, 1.3228756555322954); (2, 1.4142135623730951); (2.25, 1.5); (2.5, 1.5811388300841898); (2.75, 1.6583123951777); (3, 1.7320508075688772); (3.25, 1.8027756377319946); (3.5, 1.8708286933869707); (3.75, 1.9364916731037085); (4, 2); (4.25, 2.0615528128088303); (4.5, 2.1213203435596424); (4.75, 2.179449471770337); (5, 2.23606797749979); (5.25, 2.29128784747792); (5.5, 2.345207879911715); (5.75, 2.3979157616563596); (6, 2.449489742783178); (6.25, 2.5); (6.5, 2.5495097567963922); (6.75, 2.598076211353316); (7, 2.6457513110645907); (7.25, 2.692582403567252); (7.5, 2.7386127875258306); (7.75, 2.7838821814150108); (8, 2.8284271247461903); (8.25, 2.8722813232690143); (8.5, 2.9154759474226504); (8.75, 2.958039891549808); (9, 3). Horizontal axis: x; vertical axis: S(x). Invented teaching data.000.752.251.54.52.256.7539xS(x) S
Graph S · nonnegative-input portion · original teaching data; the table/description supplies exact coordinates.

Build your notes, then practise

Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.

Guided practice

Guided 1

For f(x) = (x − 3)² − 4, find the vertex, roots and f(0); then describe g(x) = f(x − 2) + 1.

Show a starting hint

Read each feature from the form that exposes it.

Compare with the explained answer
  1. The vertex is (3,−4); setting f(x) = 0 gives (x − 3)² = 4, so x = 1 or 5.
  2. f(0) = 9 − 4 = 5.
  3. g shifts f right 2 and up 1, so its vertex is (5,−3).

Answer / response criteria: Vertex (3,−4); roots 1 and 5; f(0) = 5; g has vertex (5,−3).

Guided 2

Compare the displayed polynomial P, rational R and radical S tables/graphs. Identify which matches x³, 4/x and √x, and explain one graph feature of each.

Compare with the explained answer
  1. P includes (−2,−8),(0,0),(2,8), matching x³ and crossing the origin.
  2. R includes (1,4),(2,2),(4,1), matching 4/x with x = 0 excluded.
  3. S includes (0,0),(1,1),(4,2),(9,3), matching √x on x ≥ 0.

Answer / response criteria: P: x³; R: 4/x; S: √x, with features as stated.

Independent practice

Independent 1

Find the input giving output 16 for f(x) = 2^x, the inputs giving 9 for g(x) = x², the input giving 3 for h(x) = √(x − 1), and the input giving 4 for r(x) = 12/x.

Compare with the explained answer
  1. 2^4 = 16, so x = 4.
  2. x² = 9 gives x = ±3.
  3. √(x − 1) = 3 gives x = 10; 12/x = 4 gives x = 3, with x ≠ 0.

Answer / response criteria: 4; −3 and 3; 10; 3.

Independent 2

Use those displays to find P(x) = −8, R(x) = 2 and S(x) = 3. State whether R has an output at x = 0 and whether S has a real output at x = −1.

Compare with the explained answer
  1. P(−2) = −8.
  2. R(2) = 2 and S(9) = 3.
  3. R excludes zero; √(−1) is not real in this domain.

Answer / response criteria: x = −2, 2 and 9 respectively; R(0) and real S(−1) are unavailable.

Application practice

Application 1

A culture follows N(t) = 200·2^(t/3). Interpret both constants, find N(6), and find when N reaches 1600. Is the model justified for unlimited time?

Compare with the explained answer
  1. There are 200 initially and one doubling every 3 time units.
  2. N(6) = 200·2² = 800.
  3. 1600/200 = 8 = 2³, so t/3 = 3 and t = 9. Unlimited growth is not established; space and nutrients may constrain it.

Answer / response criteria: Initial 200; doubling interval 3; N(6) = 800; t = 9; unrestricted extrapolation is unjustified.

Application 2

A cube model V(s) = s³ uses side length s ≥ 0. A fixed-area rectangle model L(w) = 12/w uses w > 0. Compare their graph shapes and recover s when V = 27 and w when L = 3.

Compare with the explained answer
  1. The cube model increases nonlinearly in the first quadrant.
  2. The fixed-area length decreases as positive width grows, with zero width excluded.
  3. s³ = 27 gives s = 3; 12/w = 3 gives w = 4.

Answer / response criteria: Increasing cubic versus decreasing positive reciprocal; s = 3 and w = 4.

Recall without your notes

What can you explain now?

A cube model V(s) = s³ uses side length s ≥ 0. A fixed-area rectangle model L(w) = 12/w uses w > 0. Compare their graph shapes and recover s when V = 27 and w when L = 3.

Compare with the explanation

Increasing cubic versus decreasing positive reciprocal; s = 3 and w = 4.

Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.

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Source and review status

Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.

Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.

Official source specification / guide