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Digital SAT · Math · Advanced Math

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

Core coverage follows the detailed SAT skills in Appendix B of the September 2026 framework. The conditional School Day Essay is separate. The framework is maintained rather than issued as an annual subject syllabus.

Original teaching and explained self-check practice are available. Independent subject-specialist review and difficulty calibration remain pending; this is not an exam-board endorsement or a mastery certificate.

Official source specification / guide