11.3 Nonlinear Functions
Key Takeaways
- For f(x) = ax^2 + bx + c, the sign of a tells opening: a > 0 opens up (minimum), a < 0 opens down (maximum). The y-intercept is c.
- x-intercepts come from f(x) = 0; the vertex sits on x = -b/(2a), which is also the midpoint of two x-intercepts. For x^2 - 6x + 5 the vertex is (3, -4).
- Exponential f(x) = a · b^x grows when b > 1 and decays when 0 < b < 1. A culture of 4 that doubles each hour has 4 · 2^3 = 32 cells after 3 hours.
- Absolute value y = |x - h| + k is a V with vertex (h, k). y = |x - 4| + 1 has vertex (4, 1). No trigonometry is assessed on PSAT 8/9.
- Read a graph by shape first: parabola, exponential curve, or V. Rational and radical function graphs are not PSAT 8/9 Advanced Math.
A nonlinear function is a rule whose graph is not a line. College Board's PSAT 8/9 list names quadratic, exponential, absolute value, and polynomial functions—not rational or radical functions, and not trigonometry. This independent OpenExamPrep section teaches how to read opening direction, intercepts, a vertex at grade 8/9, integer or simple-fraction exponential rates, and a V-shaped absolute-value graph. OpenExamPrep does not claim official College Board approval or exact equivalence with official materials.
If 11.2 Nonlinear Equations asked which x make an equation true, this section asks what f does as x changes and what the graph shows. Linear comparison: 9.3 Linear Functions used a constant slope. Here the rate changes. Advanced Math remains about 20% of operational Math (7–9 questions).
Quadratic functions f(x) = ax^2 + bx + c
The graph is a parabola.
| Feature | How to read it at grades 8–9 |
|---|---|
| Opening | If a > 0, the parabola opens up (a minimum). If a < 0, it opens down (a maximum). |
| y-intercept | f(0) = c, the point (0, c) |
| x-intercepts | Solve f(x) = 0, usually by factoring |
| Axis of symmetry | Vertical line x = -b/(2a), also the midpoint of two x-intercepts |
| Vertex | The turning point on the axis; its y-value is the min or max output |
The coefficient a does not only make the curve steep. Its sign is the first graph question: up or down. A larger |a| than 1 stretches vertically (narrower look); a fraction such as 1/2 makes a wider parabola. You do not need sine, cosine, or tangent to describe this curve. PSAT 8/9 does not assess trigonometry.
Example: f(x) = x^2 - 6x + 5.
a = 1 > 0, so it opens up. y-intercept: f(0) = 5, point (0, 5). Factor: x^2 - 6x + 5 = (x - 1)(x - 5). x-intercepts (1, 0) and (5, 0). Axis: midpoint of 1 and 5 is x = 3, and -b/(2a) = -(-6)/2 = 6/2 = 3. Vertex: f(3) = 9 - 18 + 5 = -4, so (3, -4). That is a minimum.
Check another point: f(2) = 4 - 12 + 5 = -3. f(4) = 16 - 24 + 5 = -3. Symmetric around x = 3.
Example: g(x) = -x^2 + 4x.
a = -1 < 0, opens down. g(x) = -x(x - 4). x-intercepts (0, 0) and (4, 0). y-intercept is also (0, 0) because c = 0. Axis: midpoint of 0 and 4 is x = 2. -b/(2a) = -4 / (2 · (-1)) = -4 / -2 = 2. g(2) = -4 + 8 = 4. Vertex (2, 4), a maximum.
Check: g(1) = -1 + 4 = 3. g(3) = -9 + 12 = 3.
Example: h(x) = 2x^2.
Opens up, vertex at (0, 0), narrower than y = x^2 because 2 > 1. h(3) = 2 · 9 = 18. Table: x = -2, -1, 0, 1, 2 maps to h = 8, 2, 0, 2, 8.
If a stem gives f(x) = -2x^2 + 3x + 1 and asks only whether the graph opens up or down, you do not need the vertex. a = -2 is enough: downward. The constant 1 is f(0), the y-intercept, not the opening cue.
Exponential growth and decay
A common 8/9 model is f(x) = a · b^x with a ≠ 0 and b > 0, b ≠ 1.
- Growth when b > 1 (repeated multiplication by more than 1).
- Decay when 0 < b < 1 (repeated multiplication by a proper fraction).
- a is the value at x = 0, because b^0 = 1, so f(0) = a.
Integer and simple-fraction bases are the ones to practice: 2, 3, 1/2, 3/2. Continuous-growth constants are not this test.
Doubling. A culture starts at 4 cells and doubles each hour: f(t) = 4 · 2^t.
| Hours t | Cells f(t) |
|---|---|
| 0 | 4 |
| 1 | 8 |
| 2 | 16 |
| 3 | 32 |
| 4 | 64 |
After 3 hours: 4 · 2^3 = 4 · 8 = 32. After 4 hours: 4 · 16 = 64. A trap is 4 · 3 = 12 (adding instead of multiplying) or 4^3 = 64 (wrong place for the 4).
Simple growth by a fraction. A collection starts at 50 stickers and is multiplied by 3/2 each week: p(w) = 50 · (3/2)^w.
p(0) = 50. p(1) = 50 · 3/2 = 75. p(2) = 50 · 9/4 = 450/4 = 112.5, which is also 75 · 3/2.
Halving (decay). A battery reading starts at 80 and halves each day: d(n) = 80 · (1/2)^n.
| Days n | Reading |
|---|---|
| 0 | 80 |
| 1 | 40 |
| 2 | 20 |
| 3 | 10 |
d(3) = 80 · 1/8 = 10.
5 · 2^x. f(0) = 5, f(1) = 10, f(2) = 20, f(3) = 40. The graph passes through (0, 5) and rises, getting steeper. It is not a parabola and not a V.
Exponential versus linear: adding 5 each time is a line; multiplying by 2 each time is exponential. A table that goes 3, 6, 12, 24 is doubling. A table that goes 3, 6, 9, 12 is +3, a line from 9.3 Linear Functions.
Absolute value: V-shape
f(x) = |x| is a V with vertex at the origin (0, 0). Left of 0 the graph falls toward 0 as x increases (slope -1 on that piece); right of 0 it rises (slope 1).
f(x) = |x - h| + k shifts the V so the vertex is (h, k).
- y = |x - 4| + 1 has vertex (4, 1).
- y = |x + 2| - 3 is |x - (-2)| - 3, vertex (-2, -3).
- y = |x| + 2 has vertex (0, 2).
- y = -|x| flips over the x-axis: an upside-down V with vertex (0, 0), opening downward.
Evaluate: if f(x) = |x - 4| + 1, then f(4) = 1, f(6) = |2| + 1 = 3, f(2) = 2 + 1 = 3, f(0) = 4 + 1 = 5. The two sides are symmetric about x = 4.
The equation |x - 3| = 5 from the previous section is the function y = |x - 3| meeting the horizontal line y = 5. The graph picture and the two-case algebra give the same two x-values, 8 and -2.
Polynomials you can evaluate
A quadratic is a degree-2 polynomial. A cubic such as p(x) = x^3 - 2x + 4 is a degree-3 polynomial. On PSAT 8/9, expect to evaluate and to compare outputs, not to run a full cubic-factorization course.
p(2) = 8 - 4 + 4 = 8. p(-1) = -1 + 2 + 4 = 5. p(0) = 4, the constant term.
q(x) = 2x^3. q(-2) = 2(-8) = -16. Odd degree with a positive leading coefficient: as x gets large positive, q grows; as x gets large negative, q is large negative. You do not need a trigonometric graph to say that.
Rational functions such as f(x) = 1/x and radical functions such as f(x) = √x are not PSAT 8/9 Advanced Math graphs. Skip them in this chapter. They appear on later SAT Suite tests.
Interpreting a graph
Read the picture before you invent algebra.
- Shape. Parabola (U), V, or exponential (flat then steep, or steep then flattening toward an axis). A straight line is Algebra, not this domain.
- Intercepts. Where it crosses the axes. Those are f(x) = 0 or f(0).
- Vertex or starting value. Lowest or highest point of a parabola; corner of a V; y-intercept of an exponential as the start.
- Direction. Opens up or down; growth or decay; upright V or flipped V.
A graph that passes through (0, 5), (1, 10), (2, 20) matches f(x) = 5 · 2^x, not f(x) = 5x + 5 (that would give (1, 10) but then (2, 15), not 20). A graph with x-intercepts 1 and 5 and a lowest point at (3, -4) matches f(x) = x^2 - 6x + 5 from the first example.
Desmos: type f(x) = x^2 - 6x + 5 and click the vertex and intercepts to confirm (3, -4), (1, 0), (5, 0), (0, 5). Use the graph as a check after you already factored on paper. More tool notes: 14.3 Reference Sheet, Desmos, and What This Test Does Not Include.
On SPR, a vertex y-value might be the typed answer: -4. Include the minus sign. Character limits and no-comma rules are in 15.1 Student-Produced Response Questions.
For f(x) = -2x^2 + 3x + 1, which statement is true?
A culture starts at 4 cells and doubles each hour. How many cells are there after 3 hours?
What is the vertex of y = |x - 4| + 1?