The most common SAT Math question type (14.0% of the bank): quadratics, exponentials, polynomials and radicals. Covers the three forms of a quadratic and what each hands you for free, the vertex as the midpoint of the zeros, exponential growth and decay as a times b to the x, percent change as a multiplier, transformations of a graph, and degree as a cap on the number of zeros — with six worked examples, four traps and three checks.
Subject: SAT Prep · 61 slides · applied lesson
Open the interactive version of this deck
Title
SAT Math · Type 1 of 19
14.0% of the question bank — 235 of 1675 questions
Objectives
A nonlinear function is any function whose graph is not a straight line: a quadratic, an exponential, a higher-degree polynomial, or something with a variable under a root. This one type is 14.0 per cent of the Math section — roughly one question in seven — and almost every version of it is answered by picking the right form and reading the answer off it.
The whole deck rests on one habit: before you compute anything, ask which form the question is already written in, and what that form hands you.
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — 235 tagged questions of this type in the site's bank
Section
Section 1
Concept
A nonlinear function is any function whose graph is not a straight line: a quadratic, an exponential, a higher-degree polynomial, or something with a variable under a root. This one type is 14.0 per cent of the Math section — roughly one question in seven — and almost every version of it is answered by picking the right form and reading the answer off it.
You will see it phrased in these ways:
Notice what is missing from that list: you are almost never asked to solve anything difficult. You are asked to read a feature off a function, and the work is choosing which form makes that feature visible.
Picture it
The same three-panel card as the survey deck, so the shorthand carries over: what identifies it, what you write first, and what is built to catch you.
Figure (svg): Nonlinear functions: the tell, the move, and the trap
Everything in this deck is an elaboration of the green panel. The rules section tells you what each form gives you; the examples show you reading it off.
Prediction
Recognition first. The test does not label these, so you have to see it.
Predict first
Three of these are nonlinear. Which one is linear?
Correct: f of x equals 3x minus 7
Why: A function is linear only when the variable appears to the first power and nowhere else — not in an exponent, not under a root, not squared. In 3x minus 7 the x is plain, so its graph is a straight line. The other three put x in an exponent, raise it to the second power, and place it under a radical respectively, and all three graph as curves. This is the whole recognition test: look at where the x is standing.
Concept
Four families sit under one label. They share a routine but differ in what you read off them.
| family | what it looks like | what you are usually asked for |
|---|---|---|
| Quadratic | an x squared term, graph is a parabola | the vertex, the zeros, or the y-intercept |
| Exponential | the x sits in the exponent | the starting value, the growth factor, or a value at a time |
| Polynomial | x to a power of 3 or more | the number of zeros, or end behaviour |
| Radical or rational | x under a root, or x in a denominator | the domain, or a value — with a check for extraneous results |
Quadratics and exponentials are the overwhelming majority. If you are short of time, learn those two families completely before touching the other two.
Definition probe
This single table is most of the type. Sort each form by the feature it makes immediately visible.
Sort into buckets
Which feature does each form give you without any work?
Discrimination
No solving. Name the family from the shape of the function alone.
Sort into buckets
Which family is each?
Warm-up
Before any rules. Try it, then check.
Discussion prompt
The function f is defined by f(x) = 5 times 3 to the power x. What is f(0)?
Hint: What is any nonzero number raised to the power zero?
Answer:
f(0) = 5. Anything to the power zero is 1, so f(0) is 5 times 1.
The general fact is worth more than the arithmetic: in a times b to the x, the number a is always the value at x equals zero — the starting amount.
The trap here is answering 15, which is 5 times 3. That is f(1), not f(0), and it is the most common wrong answer on this exact question.
Pattern
Three steps, and the first two involve no calculation at all. Most of the marks on this type are won or lost in step 2.
Name the family from where the x is standing.
Why: Squared means parabola, exponent means exponential, root means radical. This decides which set of facts applies and takes about two seconds.
Name the form it is written in, and say what that form hands you.
Why: Factored gives zeros, vertex form gives the turning point, standard gives the y-intercept, a times b to the x gives the start and the rate.
Ask whether the form you have already answers the question. Convert only if it does not.
Why: Most questions are written in the form that answers them. Expanding or re-factoring is usually the test watching you waste ninety seconds.
Step 3 is the discipline. The instinct to tidy an expression into standard form is exactly the instinct this type punishes.
Section
Section 2
Concept
Every quadratic can be written three ways, and each one is a different question already answered.
| form | written as | what it hands you |
|---|---|---|
| Standard | ax squared plus bx plus c | the y-intercept is c |
| Factored | a(x minus p)(x minus q) | the zeros are p and q |
| Vertex | a(x minus h) squared plus k | the vertex is (h, k) |
Learn this table cold. It is, on its own, worth more than any other single fact on the Math section.
Concept
A parabola is symmetric, so its turning point is exactly midway between the two points where it crosses the x-axis.
\[ x_{\text{vertex}} = \frac{p + q}{2} \]
This is the single fastest route from factored form to a vertex, and it converts a question most students expand for into one line of arithmetic.
Prediction
Factored form, and the question wants the turning point.
Predict first
The function g is defined by g(x) = (x minus 3)(x plus 5). What is the x-coordinate of the vertex?
Correct: negative 1
Why: The zeros are 3 and negative 5, read off the factors with the signs flipped. The vertex sits halfway between them, and the midpoint of 3 and negative 5 is (3 plus negative 5) divided by 2, which is negative 1. No expanding, no formula, about eight seconds of work.
Concept
An exponential model carries its two facts on its face: the value at time zero, and what happens each period.
\[ f(x) = a \cdot b^{\,x} \]
Every exponential question reduces to identifying a, identifying b, and knowing what one step of x is worth.
Prediction
The percent-to-multiplier rule, in the direction students get wrong.
Predict first
A machine loses 12 per cent of its value each year. Which base models this?
Correct: 0.88
Why: Losing 12 per cent leaves 88 per cent of the value behind, so each year you multiply by 0.88. Writing 0.12 would model keeping only 12 per cent, an 88 per cent collapse. Writing 1.12 models growth. A base is never negative in these models, because the quantity never flips sign.
Concept
Growth of r per cent means b equals 1 plus r; decay of r per cent means b equals 1 minus r.
| the story says | b is | so the model is |
|---|---|---|
| grows 5 per cent per year | 1.05 | a times 1.05 to the power t |
| falls 15 per cent per year | 0.85 | a times 0.85 to the power t |
| doubles each period | 2 | a times 2 to the power t |
| halves each period | 0.5 | a times 0.5 to the power t |
The commonest error on this rule is writing 0.15 for a 15 per cent decline. That models keeping 15 per cent, which is a 85 per cent crash.
Concept
Whatever the family, the y-intercept is what you get by substituting zero for x.
That last line catches people: the y-intercept of (x minus 3)(x plus 5) is negative 15, not negative 3 or 5.
Prediction
The rule that the intercept is f(0), applied where it is least obvious.
Predict first
What is the y-intercept of f(x) = (x minus 4)(x plus 2)?
Correct: negative 8
Why: The y-intercept is f(0). Substituting zero gives (0 minus 4) times (0 plus 2), which is negative 4 times 2, which is negative 8. The constants inside the brackets, 4 and 2, are related to the zeros rather than to the intercept, and both appear among the wrong answers for exactly that reason.
Concept
Inside the bracket, a minus moves the graph right; outside, a plus moves it up.
| change to f(x) | effect on the graph |
|---|---|
| f(x minus 3) | shifts 3 units RIGHT |
| f(x plus 3) | shifts 3 units LEFT |
| f(x) plus 3 | shifts 3 units UP |
| negative f(x) | flips vertically, across the x-axis |
| 2 times f(x) | stretches vertically, twice as tall |
The rule in one sentence: anything touching the x behaves in reverse; anything outside behaves normally.
Concept
A polynomial of degree n crosses the x-axis at most n times, and each factor names one crossing.
For a graph question, this is often enough on its own: a curve that turns twice cannot be a quadratic, because a parabola turns exactly once.
Prediction
Transformations, in the direction that feels backwards.
Predict first
The graph of y = f(x minus 6) compares to the graph of y = f(x) how?
Correct: shifted 6 units right
Why: The change is inside the function, applied to x before f acts, so it behaves in reverse: minus 6 moves the graph in the positive direction, to the right. One way to see it is to ask which input now produces the old output at x equals 0 — you need x equals 6, so every point has moved 6 to the right.
Two truths and a lie
Three of these statements about nonlinear functions are correct. Eliminate them; the one left standing is false.
Eliminate the wrong options
Which statement is FALSE?
Survives elimination: c
Why: A quadratic has AT MOST two real zeros — it can have two, exactly one (when the vertex sits on the axis), or none at all (when the parabola never reaches the axis). The discriminant decides which. This matters because how-many-solutions questions are asked directly, and the assumption that there are always two is precisely what they are testing.
Check
One question, and the whole point is which form you were handed.
Check your understanding
The function f is defined by f(x) = negative 3(x minus 4) squared plus 20. What is the maximum value of f?
Answer: A
Why: This is already in vertex form, so the vertex is (4, 20). Because a is negative 3, the parabola opens downward, which makes the vertex a maximum. The maximum value is the y-coordinate, 20. No algebra is required at all.
Note the sign flip: the bracket reads x minus 4, so h is positive 4. Had it read x plus 4, the vertex would sit at negative 4.
Section
Section 3
Worked example
The function f is defined by f(x) = (x minus 2)(x minus 8). What is the minimum value of f?
Figure (svg): A parabola crossing the x-axis at 2 and 8 with its vertex marked at (5, negative 9)
Read the zeros off the factors: x equals 2 and x equals 8.
Why: A product is zero when either factor is zero, and the sign flips from what is written.
Average them to find the vertex x-coordinate: (2 plus 8) divided by 2, which is 5.
Why: The parabola is symmetric, so the turning point sits midway between the crossings.
Substitute x equals 5: f(5) = (5 minus 2)(5 minus 8) = 3 times negative 3.
Why: The minimum VALUE is the y-coordinate, so you must evaluate rather than stop at x equals 5.
The trap answer is 5 — the x-coordinate. The question asked for the minimum value, which is a y-coordinate, and 5 will be sitting among the choices.
Verify: test a nearby point, f(4) = 2 times negative 4 = negative 8.
Why: Negative 8 is greater than negative 9, which is consistent with negative 9 being the minimum.
Answer: negative 9
Worked example
A town of 8,000 people grows by 3 per cent per year. Which function gives the population after t years?
Figure (svg): An exponential growth curve starting at 8000 and rising by 3 per cent a year
Identify a, the starting value: 8,000.
Why: a is the value at t equals 0, and the story gives the population before any growth.
Convert 3 per cent growth into a base: 1 plus 0.03, which is 1.03.
Why: Growth of r per cent means multiplying by 1 plus r each period.
Write the model with t counting years.
Why: The exponent must count the same periods the rate is quoted in.
Wrong answers will offer 0.03 as the base, 1.3 instead of 1.03, and 8000 times 3 to the power t. All three are decimal-placement errors rather than conceptual ones.
Verify: check t equals 1 gives 8,000 times 1.03 = 8,240, a rise of 240.
Why: 240 is 3 per cent of 8,000, so the model reproduces the stated rate.
Answer: P(t) = 8000 times 1.03 to the power t
Worked example
A substance decays so that 60 per cent remains after each hour. What percentage is lost each hour, and what is the base?
Figure (svg): Bars shrinking from 100 to 60 to 36 to 21.6 as 60 per cent survives each hour
The base is what remains, so b equals 0.60.
Why: The multiplier is always the fraction that survives, not the fraction lost.
The amount lost is the complement: 1 minus 0.60, which is 0.40.
Why: What is left and what is lost must sum to the whole.
So 40 per cent is lost each hour.
Why: Converting the decimal back to a percentage answers the question as asked.
This example runs the percent rule in the direction the test prefers: it gives you the multiplier and asks for the percentage, rather than the other way round.
Verify: check with 100 units: after one hour 60 remain, so 40 were lost.
Why: 40 out of 100 is 40 per cent, matching the answer.
Answer: 40 per cent is lost; the base is 0.60
Step zero
Not the solution. Only the opening line.
Discussion prompt
A question reads: the function h is defined by h(x) = negative 2(x plus 4) squared plus 18. What is the maximum value of h? What do you write first, and why do you not need to do any algebra?
Hint: Which form is this already in, and what does that form hand you?
Answer:
You write down the vertex: (negative 4, 18). The function is already in vertex form, a(x minus h) squared plus k, so the turning point is visible.
The sign of h flips: the bracket reads x plus 4, which means h is negative 4.
Because a is negative 2, the parabola opens downward, so the turning point is a maximum rather than a minimum.
The maximum value is the y-coordinate, 18. There is no algebra to do at all — and the trap answer is negative 4, the x-coordinate.
Worked example
The graph of a polynomial function turns three times. What is the smallest possible degree?
Figure (svg): A curve with three marked turning points
Count the turns: three.
Why: Each turn is a change of direction, from rising to falling or the reverse.
A polynomial of degree n turns at most n minus 1 times.
Why: A parabola, degree 2, turns once; a cubic, degree 3, turns at most twice.
Solve n minus 1 is at least 3, so n is at least 4.
Why: Three turns require the degree to be one more than the number of turns, at minimum.
Graph-shape questions are answered by counting turns, not by trying to find an equation. Turns plus one is the minimum degree.
Verify: check degree 3 would allow only two turns.
Why: Two is fewer than the three turns observed, so degree 3 is impossible and 4 is the smallest that works.
Answer: degree 4
Worked example
The point (2, 5) lies on the graph of y = f(x). Which point must lie on the graph of y = f(x minus 3) plus 4?
Figure (svg): The point (2, 5) moved by an arrow to (5, 9)
The inside change, minus 3, shifts the graph 3 units right.
Why: Changes to x act in reverse, so minus moves in the positive direction.
The outside change, plus 4, shifts the graph 4 units up.
Why: Changes outside the function behave normally.
Apply both to the point: 2 plus 3 gives 5, and 5 plus 4 gives 9.
Why: The x-coordinate takes the horizontal shift and the y-coordinate takes the vertical one.
The substitution in the verify line is the safest method of all. If you ever doubt which way a shift goes, ask which input reproduces the old one.
Verify: check by substitution: at x equals 5, f(5 minus 3) plus 4 = f(2) plus 4 = 5 plus 4 = 9.
Why: The new function does evaluate to 9 at x equals 5, confirming the point.
Answer: (5, 9)
Faded example
From memory. These are the three lines that carry most of this question type.
Fill in the blanks
Standard form ends in the constant c, which is the y-intercept. Factored form a(x minus p)(x minus q) hands you the zeros. Vertex form a(x minus h) squared plus k hands you the vertex, at the point (h, k). And the vertex always sits halfway between the two zeros.
Why: Those four facts answer the large majority of quadratic questions on the test without any manipulation. The recurring theme is that the sign of p, q and h flips between what is written in the bracket and what it means on the graph.
Worked example
For f(x) = x squared minus 10x plus 21, find the vertex — without the vertex formula.
Figure (svg): A parabola crossing at 3 and 7 with its vertex at (5, negative 4)
Try factoring first: two numbers multiplying to 21 and adding to negative 10 are negative 3 and negative 7.
Why: Factoring is cheaper than completing the square when the numbers are friendly.
So f(x) = (x minus 3)(x minus 7), and the zeros are 3 and 7.
Why: Factored form makes the zeros immediately visible.
Average the zeros: (3 plus 7) divided by 2, which is 5. Then f(5) = 25 minus 50 plus 21.
Why: The vertex sits midway between the zeros, and its height needs one substitution.
This is the whole deck in one example: the question arrived in standard form, and the fastest path was to move to the form that answers it.
Verify: check the arithmetic: 25 minus 50 is negative 25, plus 21 is negative 4.
Why: The vertex is (5, negative 4), and testing f(4) gives 16 minus 40 plus 21 = negative 3, which is higher, as it must be.
Answer: the vertex is (5, negative 4)
Fill the middle
The percent-to-multiplier conversion, in both directions.
Fill in the blanks
A value that grows 7 per cent per year has base 1.07. A value that falls 7 per cent per year has base 0.93. A value that triples each period has base 3.
Why: Growth adds to 1 and decay subtracts from it, which is why a 7 per cent fall gives 0.93 rather than 0.07. Tripling is a whole multiplier, so the base is simply 3. Getting 0.07 for the decay case is the single most common error on exponential modelling questions.
Estimation
Exponential growth is much faster than intuition suggests, and estimating protects you from choosing an answer that is off by an order of magnitude.
Predict first
A colony of 100 bacteria doubles every hour. Roughly how many are there after 10 hours?
Correct: about 100,000
Why: Doubling ten times multiplies by 2 to the power 10, which is 1,024 — a useful number to know, since it is a bit over one thousand. So 100 becomes about 102,400. The tempting wrong answer is 1,000, which comes from multiplying 100 by 10 and treating the growth as linear. Exponential growth compounds; it does not add.
Check
A story to convert. Watch the base.
Check your understanding
An investment of 5,000 dollars earns 4 per cent interest per year, compounded annually. Which function gives its value after t years?
Answer: A
Why: The starting value is 5,000, so a is 5000. Growth of 4 per cent means multiplying by 1 plus 0.04, so the base is 1.04, and t counts years. Checking after one year gives 5,000 times 1.04, which is 5,200 — a gain of 200, and 200 is indeed 4 per cent of 5,000.
Section
Section 4
Trap
The trap. You are handed f(x) = (x minus 2)(x minus 8) and asked for the vertex.
You multiply it out to x squared minus 10x plus 16, then reach for the vertex formula, negative b over 2a, and grind through it. Ninety seconds gone, and two opportunities for an arithmetic slip introduced.
The factored form you started with already contained the answer.
The fix. Before manipulating anything, ask what the current form already gives you.
The zeros were visible: 2 and 8. The vertex is halfway between them, at 5. That is one line of arithmetic.
Trap
The trap. The function is f(x) = 3(x plus 5) squared minus 2, and you report the vertex as (5, negative 2).
The bracket reads x plus 5, and vertex form is written a(x minus h) squared plus k. So x plus 5 is really x minus (negative 5), which makes h equal negative 5.
The same flip catches people on zeros: the factor (x plus 3) gives a zero at x equals negative 3, not positive 3.
The fix. Read the form as a subtraction always, and ask what number is being subtracted.
In (x plus 5), what is subtracted is negative 5. In (x minus 7), what is subtracted is 7.
Error analysis
A student's work on a vertex question. One line is wrong. Find it before reading the notes.
Annotate
On: \( f(x) = (x - 2)(x - 8) \;\Rightarrow\; \text{vertex } x = \frac{2 + 8}{2} = 5 \;\Rightarrow\; \text{minimum} = 5 \)
The most expensive errors on this type are not algebra errors. They are answering a slightly different question, perfectly.
Trap
The trap. The question asks for the minimum value of a function, and you find that the vertex is at x equals 5, so you answer 5.
The value of a function is a y-coordinate. x equals 5 is where the minimum happens, not what it is.
This is the same trap as answering x when the question wanted 2x, and it is offered on this type more than on any other.
The fix. Decide, before solving, whether the question wants a location or a value.
Minimum value, maximum value, and the value of f both mean a y-coordinate. Where the minimum occurs and the axis of symmetry mean an x-coordinate.
Elimination
A car worth 20,000 dollars loses 15 per cent of its value each year.
Eliminate the wrong options
Which function models its value after t years? Three can be ruled out on structure alone.
Survives elimination: b
Why: Losing 15 per cent leaves 85 per cent, so the base is 0.85 and the model is 20,000 times 0.85 to the power t. Each wrong answer is a distinct, predictable confusion: using the percentage lost as the base, using growth instead of decay, and treating a percentage change as a fixed subtraction. Recognising those three patterns lets you eliminate without arithmetic.
Trap
The trap. A population grows 10 per cent per year, and after 3 years you compute a 30 per cent increase.
Percentages do not add across periods; multipliers multiply. Three years of 10 per cent growth is 1.10 times 1.10 times 1.10, which is about 1.331 — a 33.1 per cent rise, not 30.
Over longer spans the gap widens sharply: ten years of 10 per cent growth is a 159 per cent rise, not 100 per cent.
The fix. As soon as you see a repeated percentage, write it as a base raised to a power.
The whole point of an exponential model is that the change is applied to a quantity that has already changed.
Counterexample
A claim that sounds right and is not.
Discussion prompt
A student says: every parabola crosses the x-axis twice, so every quadratic has two zeros. Give a counterexample, and say what decides the real answer.
Hint: Think about where the vertex sits relative to the axis.
Answer:
Counterexample: f(x) = x squared plus 1. Its vertex is at (0, 1), above the x-axis, and it opens upward — so the curve never reaches the axis and there are no real zeros.
A second counterexample: f(x) = x squared. Its vertex sits exactly on the axis, giving exactly one zero.
So a quadratic has two, one, or no real zeros, and what decides it is the discriminant, b squared minus 4ac: positive gives two, zero gives one, negative gives none.
The geometric version is easier to remember: it depends on whether the vertex is below, on, or above the axis — and which way the parabola opens.
Edge cases
The vertex-is-halfway-between-the-zeros rule is reliable. Where does it stop being usable?
Discussion prompt
When can you NOT find a vertex by averaging the zeros, and what do you do instead?
Hint: The rule needs something to average.
Answer:
It fails when there are no real zeros — if the parabola never crosses the axis, there is nothing to average.
In that case use completing the square to reach vertex form, or the formula x equals negative b over 2a.
The rule still works fine when there is exactly one zero: the vertex sits on it, and averaging a number with itself returns that number.
It also works when the zeros are ugly or irrational — averaging is still valid, it is just less pleasant than reading vertex form directly.
The practical guidance: try factoring first, and fall back to negative b over 2a only when factoring fails. On the SAT the numbers are usually chosen so that factoring works.
Check
Two rules at once: factored form, and what the question actually asked for.
Check your understanding
The function g is defined by g(x) = 2(x plus 1)(x minus 7). At what value of x does g attain its minimum?
Answer: A
Why: The zeros are negative 1 and 7, read off the factors with the signs flipped. The vertex sits halfway between them: (negative 1 plus 7) divided by 2, which is 3. Since a is positive 2, the parabola opens upward and the vertex is a minimum. The question asked at what value of x, so the answer is the x-coordinate, 3.
Note that this check reverses the previous trap: here the question wanted the x-coordinate, and the y-coordinate was the distractor. Read the stem, every time.
Section
Section 5
Matching
The core table of this deck, as a matching exercise. No calculation.
Match the pairs
Why: The first three rows are the same quadratic written three ways — expand any of them and you get the others. That is the point of the type: one function, three presentations, and each presentation answers a different question with no work at all.
Sorting
The decision that step 2 of the routine asks for.
Sort into buckets
Which form does each question want?
Three of the six wanted vertex form. That is a fair reflection of the real balance — the vertex is the single most-asked-for feature on this type.
Comparison
Fill the blanks from memory. If you can complete this table you have the core of the type.
Comparison matrix
| form | gives you free | costs you | when to convert to it |
|---|---|---|---|
| Standard | the y-intercept | vertex and zeros both hidden | when asked for the value at x equals 0 |
| Factored | the zeros | only exists if it factors | when asked where it crosses, or how many roots |
| Vertex | the turning point (h, k) | needs completing the square | when asked for a maximum or minimum |
The costs column is the one students never think about. Factored form is the cheapest to obtain when the numbers are friendly, and impossible when they are not — which is why factoring is the first thing to try and never the thing to insist on.
Trade off
There is more than one way to find a turning point. Fill in what each costs.
Comparison matrix
| method | when it is fastest | when it fails |
|---|---|---|
| Average the zeros | when the function is already factored | when there are no real zeros |
| Negative b over 2a | when it is in standard form and will not factor | never fails, but needs a substitution for the y-value |
| Read vertex form directly | when it is already in vertex form | otherwise needs completing the square |
| Graph it in Desmos | when the numbers are ugly | slower than reading a form you already have |
All four are correct methods. The skill being tested is not which one you can execute, but which one you reach for — and the answer is almost always the one that exploits the form you were handed.
Real world
One minute of thinking about why this type exists.
Discussion prompt
Compound interest, population growth, radioactive decay and drug half-lives are all the same function. What makes them exponential rather than linear, and why does that matter for the questions the SAT asks?
Answer:
They are exponential because the change each period is a proportion of the current amount, not a fixed quantity. Interest is paid on the balance you have now, including interest already earned.
Linear change adds the same number every period; exponential change multiplies by the same number every period.
That is exactly why the SAT keeps offering a linear distractor such as 5000 plus 200t alongside the correct 5000 times 1.04 to the power t. Over one period they agree; over ten they diverge sharply.
The practical tell in a word problem: per cent of signals exponential; per year attached to a fixed amount in dollars or units signals linear.
Ranking
All four start at 100. Order them from smallest to largest after 10 periods.
Put in order
Why: Decay reaches 100 times 0.9 to the power 10, about 35. The linear one adds 10 ten times, reaching exactly 200. Growth of 10 per cent per period reaches 100 times 1.1 to the power 10, about 259 — it starts slower than the linear one but overtakes it before the tenth period. Doubling reaches 100 times 2 to the power 10, which is 102,400. Two lessons: compounding beats adding eventually but not immediately, and a base of 2 dwarfs a base of 1.1 completely.
Concept
This type is 14.0 per cent of the section, so it deserves the largest single block of your Math preparation. Four sessions, in this order.
| session | what you do | why |
|---|---|---|
| 1 | Drill the three forms only. Twenty questions, all quadratics, all asking for a vertex, a zero or an intercept. | You are building the reflex that names the form before doing anything else. |
| 2 | Exponentials only. Twenty story problems, converting percentages and doubling times into bases. | The percent-to-multiplier conversion is the highest-frequency error in the whole family. |
| 3 | Mixed quadratics and exponentials, with the type label hidden. | Recognition only develops under mixing; drilling one family at a time teaches you nothing about spotting it. |
| 4 | Twenty questions under time, then log every miss as tell, move or trap. | Speed on this type frees the minutes you need for the rest of the section. |
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — filter the bank to this skill tag — 235 questions, roughly a third of each difficulty
Explain it to yourself
Close the deck. Retrieval, not recognition.
Discussion prompt
Without looking, name the three forms of a quadratic, say what each one hands you for free, and state where the vertex sits relative to the zeros.
Hint: Two of the three forms flip a sign when you read the answer off them.
Answer:
Standard, ax squared plus bx plus c — the y-intercept is c.
Factored, a(x minus p)(x minus q) — the zeros are p and q, with the signs flipped from the brackets.
Vertex, a(x minus h) squared plus k — the vertex is (h, k), again with the sign of h flipped.
The vertex sits halfway between the zeros, so averaging them gives its x-coordinate.
If you produced all four lines, you can answer most quadratic questions on the test without algebra.
Explain it
Two minutes, out loud, to someone who keeps getting it wrong.
Discussion prompt
A friend keeps writing 0.15 as the base for something that falls 15 per cent per year. How do you explain the mistake so it does not come back?
Answer:
Ask them what is left after the fall. If you lose 15 per cent, you keep 85 per cent — so the multiplier is 0.85.
Then show what 0.15 actually models: start with 100, multiply by 0.15, and you get 15. That is not a 15 per cent loss, it is an 85 per cent loss.
Give them the sentence to say every time: the base is what survives, not what disappears.
Finally, connect it to growth so the pair is symmetric: growth of 15 per cent keeps everything and adds more, so the base is 1.15. Decay keeps less than everything, so the base is below 1 but above 0.
Commit first
Commit before you check.
Predict first
The function f is defined by f(x) = (x plus 6)(x minus 2). What is the x-coordinate of its vertex?
Correct: negative 2
Why: The zeros are negative 6 and positive 2, read off the brackets with the signs flipped. Their midpoint is (negative 6 plus 2) divided by 2, which is negative 4 divided by 2, or negative 2. The tempting wrong answer is negative 4, which is the sum of the zeros without dividing by two — an easy slip to make when working quickly.
Connect it up
Blank paper. This is the single best revision artefact you can make for this type.
Draw it
Draw a map of nonlinear functions. Put the four families across the top: quadratic, exponential, polynomial, radical. Under quadratic, draw the three forms and label what each hands you for free. Under exponential, write a times b to the x and label a and b. Add the vertex-is-halfway-between-the-zeros rule with a small sketch of a parabola. Finally, list the four traps down one side: expanding unnecessarily, the sign inside the bracket, answering the x-coordinate, and treating growth as linear.
Exit ticket
One question before you close the deck.
Predict first
You are handed f(x) = (x minus 1)(x minus 9) and asked for the minimum value. What is the very first thing you should do?
Correct: Read the zeros off the brackets
Why: The function is already factored, so the zeros, 1 and 9, are visible at no cost. Averaging them gives the vertex at x equals 5, and one substitution gives the minimum value of negative 16. Expanding, using negative b over 2a, and completing the square all reach the same answer, but each begins by destroying the form that already contained what you needed.
Recap
One type, one habit: name the form, then read the answer off it.
| never do this | do this instead |
|---|---|
| Expand a factored quadratic to find its vertex | Average the zeros you were already given |
| Read (x plus 5) as a vertex at positive 5 | Flip the sign: it is negative 5 |
| Use 0.15 as the base for a 15 per cent decline | Use 0.85 — the base is what survives |
| Multiply the rate by the number of years | Raise the multiplier to the power of the years |
| Answer the x-coordinate when asked for a value | Substitute back to get the y-coordinate |
| Assume a quadratic always has two zeros | It has at most two; check the discriminant |
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — and the site's SAT pages to drill this type in isolation, then mixed
Want this taught 1-on-1? Alexander tutors SAT Prep — $55/session, free consultation.