Nonlinear functions

This skill point covers every function on the exam whose rate of change is not constant: quadratics above all, then exponentials, then polynomials of higher degree, with the occasional rational or radical expression at the edges. The tasks recur in four shapes — evaluate or interpret a nonlinear model, rewrite it into an equivalent form, read a feature (vertex, intercepts, end behaviour) off a graph or an equation, and build a model from a description.

College Board publishes the domain weight: Advanced Math is about 35% of the Math section, roughly 13 to 15 of the 44 questions, tied with Algebra as the largest domain. What College Board does not publish is how those questions split among its three skill points, or how they distribute between the first module and the harder second module. The widely repeated claim that Advanced Math dominates the hard module is a prep-industry pattern claim reverse-engineered from released material — plausible, consistent with what the released tests look like, and not an official figure. Plan with it; don't cite it as fact.

The thing being tested is almost never the algebra. Completing the square is four lines. Factoring a quadratic is one. What separates a right answer from a wrong one is whether you noticed which form of the same function makes the question trivial before you started writing. There is a weak path and a strong path through nearly every item on this skill, and they differ by about two minutes.

Weak path: an item gives you h(t) = −16t² + 64t + 80 and asks for the maximum height, and you reach for the quadratic formula, get two ugly roots, and stall. Strong path: you notice that a maximum is a vertex, that a vertex is halfway between the roots, that the roots come out of factoring in one line, and you are done in twenty seconds. Same knowledge, different order of operations, three-minute swing on a section where three minutes is roughly two questions.

If you can look at f(x) = 3(x − 2)² − 5 and say "opens up, vertex at (2, −5), minimum value −5" without pausing, skip ahead. If any part of that sentence is unfamiliar, this is the highest-value block on the page, because these objects reappear in every other Advanced Math skill.

A linear function adds the same amount at every step: 5, 9, 13, 17 — always plus 4. Nonlinear means the amount added is not constant. Two patterns cover almost everything the exam uses. A QUADRATIC adds an amount that itself grows by a fixed amount: 1, 4, 9, 16 — the jumps are 3, 5, 7, and the jumps between those jumps are 2, 2. An EXPONENTIAL multiplies by the same factor at every step: 5, 15, 45, 135 — always times 3. Addition of a constant is linear; addition of a growing constant is quadratic; multiplication by a constant is exponential. That is the entire taxonomy.

A quadratic is any function you can write as f(x) = ax² + bx + c with a ≠ 0, and its graph is a parabola: a symmetric U-shape. If a is positive the U opens upward and its lowest point is the VERTEX; if a is negative the shape opens downward, ∩, and the vertex is its highest point. The vertical line through the vertex is the AXIS OF SYMMETRY, and the graph on one side of it is a mirror image of the graph on the other.

Four words for one thing, and the exam switches between them inside a single question: a ZERO of f, a ROOT of the equation f(x) = 0, an x-INTERCEPT of the graph, and a SOLUTION of f(x) = 0 are all the same object — an input that makes the output zero. Separately, f(0) is the y-INTERCEPT, also called the initial value or the starting amount. Zeros are inputs; the y-intercept is an output. Confusing the two is the most expensive vocabulary error on the Math section.

The exponent rules you actually need, and there are only five. b^0 = 1 for any nonzero b. b^(−n) = 1/b^n. b^(1/2) = √b, and more generally b^(m/n) is the n-th root of b^m. b^m · b^n = b^(m+n). (b^m)^n = b^(mn). That last one is what lets you rewrite 9^x as 3^(2x) or 8^(x/3) as 2^x, which the exam asks for directly.

An exponential function is f(x) = a·b^x with a ≠ 0 and b > 0. Its graph either rises slowly and then steeply (when b > 1, growth) or falls steeply and then flattens out (when 0 < b < 1, decay). The flattening never finishes: a positive exponential approaches zero but never reaches it, which is why "in what year does the population reach 0" has no answer for such a model and the exam will ask when it falls below some level instead.

A POLYNOMIAL is a sum of terms of the form (number)·x^(whole number): 4x³ − 2x + 7 is one, and so is every linear and quadratic function. Its DEGREE is the largest exponent. Degree is what decides the coarse shape of the graph and what its two ends do, which is the last teach section on this page.

The same parabola can be written three ways. They are algebraically identical — expand any of them and you get the others — but each one displays a different feature for free and hides the rest. Choosing the form is the skill; converting between them is the labour.

STANDARD FORM, f(x) = ax² + bx + c. Free: the y-intercept is c, because f(0) = c. Free: the sign of a tells you which way it opens. One step away: the axis of symmetry is x = −b/(2a), and substituting that back gives the vertex's y-coordinate. Use standard form when the question asks about the y-intercept, or hands you a table of values, or asks about the discriminant.

FACTORED FORM, f(x) = a(x − r)(x − s). Free: the x-intercepts are r and s, because a product is zero exactly when one of its factors is zero. Free: the axis of symmetry is the midpoint of the roots, x = (r + s)/2 — no formula needed, just symmetry. Use factored form for anything about zeros, ground level, break-even, "when does it equal zero", or where the maximum happens. Watch the signs: (x − 3)(x + 7) has roots +3 and −7.

VERTEX FORM, f(x) = a(x − h)² + k. Free: the vertex is (h, k), the axis of symmetry is x = h, and the maximum or minimum VALUE of the function is k. Use it for any question containing the words maximum, minimum, greatest, least, highest, or "the value at which the model peaks". The h is subtracted by the form itself, so a plus sign inside means a negative h: 2(x + 5)² − 3 has vertex (−5, −3).

Converting, in the direction you will actually need. Vertex or factored to standard: expand. Standard to factored: factor, or find the roots with the quadratic formula x = (−b ± √(b² − 4ac))/(2a) and rebuild. Standard to vertex: complete the square, which always works. The recipe, with the step people drop marked — 1) factor a out of the x² and x terms ONLY, leaving the constant outside; 2) halve the coefficient of x inside the parentheses and square it; 3) add and subtract that square inside; 4) write the perfect square and carry the subtracted piece out, MULTIPLYING IT BY a on the way; 5) combine with the constant. Step 4 is where most of the errors on this conversion live.

There is also a shortcut that skips completing the square entirely, and it is usually faster: find the axis of symmetry with x = −b/(2a), then substitute that value back into the original equation to get k. Two lines, no square to complete, and it works even when a is ugly. Complete the square when the item asks for the equation in vertex form; use −b/(2a) when it only asks for the vertex or the extreme value.

The DISCRIMINANT, b² − 4ac, is the piece under the radical in the quadratic formula, and on its own it answers a whole family of items. Positive: two distinct real solutions, so the graph crosses the x-axis twice. Zero: exactly one real solution, so the vertex sits on the x-axis and the quadratic is a perfect square. Negative: no real solutions, so the graph never touches the axis. Any item containing "exactly one solution", "no real solutions", "two distinct x-intercepts", or "intersects at exactly one point" is a discriminant item, and it wants b² − 4ac set equal to, greater than, or less than zero — not solved.

f(x) = a·b^x has exactly two parameters and they do exactly two jobs. a is the output when the input is zero — the initial amount, the starting population, the purchase price. b is the number the output is multiplied by for each one-unit increase in the input. Everything an exponential item asks about is one of those two numbers, read correctly.

The translation that decides the most points: b = 1 + r, where r is the rate of change written as a decimal. Growing by 8% means multiplying by 1.08. Falling by 15% means multiplying by 0.85, because keeping 85% is what losing 15% means. Doubling is b = 2, which is a 100% increase. Halving is b = 0.5, a 50% decrease. Rising by 150% is b = 2.5, not 1.5. Read a base of 1.03 as "+3%" and a base of 0.97 as "−3%" on sight, and never read the base itself as the percentage.

Growth versus decay is a single comparison: b > 1 is growth, 0 < b < 1 is decay, b = 1 is a constant. The base in a real-world model is always positive, so a negative sign in front of a model belongs to a, not b.

Now the part that separates hard items from easy ones. When the multiplier applies once per period and the input is measured in some other unit, the model is f(t) = a·b^(t/p), where p is the length of one period in the input's units. The exponent is not a time — it is a COUNT of how many multiplications have happened. A sample that loses 12% every 5 years, with t in years, is 800(0.88)^(t/5). A culture that doubles every 6 hours, with t in hours, is a·2^(t/6). A drug with a 30-day half-life, with d in days, is a·(0.5)^(d/30). Ask "how many periods have gone by?" and the exponent writes itself.

Spotting an exponential in a description: the giveaway is a constant PERCENT or a constant FACTOR per step — "decreases by 4% annually", "doubles every decade", "triples each hour", "loses a third of its charge every 8 hours". A constant AMOUNT per step — "falls by 40 units a year", "$13.50 per poster" — is linear, and none of the machinery on this page applies to it. If you cannot name a fixed number being added each step, it is not linear.

Spotting one in a table with equally spaced inputs: divide each output by the one before it. Constant ratio means exponential. Constant difference means linear. Constant SECOND difference — the differences of the differences — means quadratic. Three divisions or three subtractions settle it, and the exam does build tables that reward the check.

One convention worth knowing because it looks like a contradiction: compound-interest items are written a(1 + r/n)^(nt), where r is the stated annual rate and n is the number of compounding periods per year. That is a model the item stipulates, not a re-expression of a true annual multiplier, and the two give slightly different numbers. Read whichever the item defines and don't import the other.

Transformations follow one rule with one twist. Changes OUTSIDE the function act on the outputs and do what they look like: f(x) + k moves the graph up k, f(x) − k moves it down, c·f(x) stretches it vertically by a factor of c, and −f(x) flips it over the x-axis. Changes INSIDE the parentheses act on the inputs and run BACKWARDS: f(x − h) moves the graph RIGHT h, f(x + h) moves it LEFT h, and f(−x) flips it over the y-axis.

The reason the inside runs backwards is worth ten seconds: to make g(x) = f(x − 3) produce the output f used to produce at 5, you now have to feed in 8. Every output arrives three units later, so the picture slides right. This is also why vertex form is readable at all — a(x − h)² + k is nothing but y = x², moved right h, up k, and stretched by a, with the transformation written into the notation.

The same rules apply to exponentials, and they add one feature. a·b^x + d has a horizontal asymptote at y = d: the outputs approach d and never arrive. With d = 0 that means the model never reaches zero, no matter how long you wait. Items exploit this by asking what value the model "approaches" or by ruling out an answer choice that has the quantity hitting zero.

END BEHAVIOUR is what the outputs do as x runs far to the left and far to the right, and for a polynomial it is decided by the leading term alone — the term with the highest power. Every other term is eventually negligible next to it. Two facts settle every case. Parity of the degree: an EVEN degree sends both ends the same way; an ODD degree sends them opposite ways. Sign of the leading coefficient: positive sends the right-hand end UP, negative sends the right-hand end DOWN. Fix the right end from the sign, then let the parity tell you the left.

Reading the leading term off factored form without expanding: the degree is the sum of the exponents on the factors, and the leading coefficient is the constant out front times the leading coefficients inside. So −2(x − 4)²(x + 1)³ has degree 2 + 3 = 5 and leading term −2x⁵ — odd degree, negative coefficient, therefore up on the left and down on the right. Multiplying it all out to learn that would cost two minutes and buy nothing.

At this point the three forms are automatic and the multiplier is second nature. Six things are still moving points: multiplicity and end behaviour on higher-degree polynomials, discriminant items where the unknown is a coefficient, rewriting an exponential over a different base or period, tangency between a line and a curve, quadratics with parameters instead of numbers, and a short list of execution errors that survive knowing every rule above.

MULTIPLICITY. In factored form, the exponent on a factor decides what the graph does at that zero. An ODD exponent means the graph CROSSES the axis there — exponent 1 crosses straight through, exponent 3 crosses with a visible flattening. An EVEN exponent means the graph TOUCHES the axis and turns back without changing sign. So −2(x − 4)²(x + 1)³ crosses at x = −1 and merely touches at x = 4. The default assumption that every zero is a crossing is what makes this item type work, and "which could be the graph" questions are usually decided by exactly one even-multiplicity factor.

DISCRIMINANT ITEMS WITH AN UNKNOWN. When a coefficient is a letter and the item states how many solutions there are, the discriminant becomes an equation or an inequality in that letter. "x² + kx + 9 = 0 has exactly one real solution" gives k² − 36 = 0, so k = ±6 and the item will ask for the positive one, the negative one, or their sum — read which. "No real solutions" gives k² − 36 < 0, so −6 < k < 6, an interval and not a value. Two answers are the norm here, and choosing the wrong one of two correct roots is a pure reading error.

TANGENCY is a discriminant item in disguise. A line and a parabola meet at exactly one point when substituting one into the other produces a quadratic with discriminant zero. For y = x² + 3 and y = mx: x² − mx + 3 = 0, so m² − 12 = 0 and m = ±2√3. Nothing about the geometry needs to be visualised — "exactly one intersection point" and "exactly one solution" are the same sentence.

REWRITING EXPONENTIALS. Two moves cover it. Change of base uses (b^m)^n = b^(mn): 9^x = 3^(2x), 8^(x/3) = 2^x, 4^(x+1) = 4·4^x. Change of period uses the same identity in reverse: 3,000(1.06)^t with t in years is identical to 3,000(1.06^(1/12))^(12t) with the exponent in months — the base and the exponent change TOGETHER, never one alone. This is also where the most-missed inference lives: a quantity that doubles every 7 years grows by 2^(1/7) ≈ 1.104 per year, about 10.4% annually, not 100/7 ≈ 14.3%. Compounding and division are not the same operation.

PARAMETERS INSTEAD OF NUMBERS. Two identities are worth carrying because they answer items without solving: for ax² + bx + c = 0, the sum of the solutions is −b/a and their product is c/a. "What is the sum of the solutions of 2x² − 14x + 9 = 0" is 14/2 = 7 in one line, and the roots themselves are irrational. Likewise, second differences of a quadratic sampled at unit-spaced inputs are constant and equal 2a, which turns a table into a leading coefficient without any algebra.

CONTEXTUAL DOMAIN, which the exam uses to punish an otherwise perfect solution. A projectile model has two zeros and only one of them is a time. A revenue model peaks and then predicts negative revenue. A population model that decays never reaches zero, so a question phrased "reaches zero" is either about a different model or wants the first time it falls BELOW a threshold. Read the situation back into the answer before selecting it: a negative length, a negative count, or a time before launch is a signal, not an answer.

EXECUTION ERRORS, which is where the remaining points actually live. Losing the a when distributing out of a completed square. Reporting h when the question asked for k. Answering with the time when the question asked for the height, or with the height when it asked for the time — both numbers are sitting in your working, and both are in the answer choices. Rounding an exponential intermediate: 0.88^4 carried as 0.60 instead of 0.5997 misses a student-produced response. Taking only the positive root when the question says "least possible value". And the format rule, which is College Board's own: student-produced responses reject mixed numbers, so two and a half is entered as 5/2 or 2.5 and never as 2 1/2, and a repeating decimal must fill the entire entry field — 2/3 entered as .66 is marked wrong, while .6666 or .6667 is accepted. The negative sign counts against the character budget.