SAT Nonlinear Functions: Every Question Type and Method

A nonlinear function's graph isn't a line: its outputs don't change by the same amount for each step in $x$. On the digital SAT, nonlinear functions is one of the three skills in Advanced Math, which makes up about 35% of the Math section, or 13 to 15 of its 40 scored questions. Its questions come in four families:

• Quadratic functions: parabolas, their vertex and zeros, and the forms that show them.
• Exponential functions: growth and decay by the same factor, like “increases by 15% every 4 days”.
• Polynomial, rational and radical functions: evaluating them, reading their graphs and their asymptotes.
• Transformations: a known graph shifted, reflected or stretched into a new one.

This guide sorts them into six groups by what the question asks you to do. The other two Advanced Math skills are in SAT Advanced Algebra. The table below gives each one's method, and each section after it teaches the method, says when Desmos is faster, and gives you an SAT-style question to try. On most of these questions, finding the number is the easy part: you earn the point by knowing which number the question wants.

How to answer an SAT nonlinear functions question

The same four moves work on every question in this guide:

1. Read the last line first. Is it asking for an input, an output, a whole point, an interval, or what a number means?
2. Translate the words into a fact about $x$ and $f(x)$: a zero means $f(x)=0$, a $y$-intercept means $x=0$, a maximum means the greatest output.
3. Pick the faster route. When the form already shows the number, read it. When the equation hides it, or a line might cross the graph more than once, graph it in Desmos and click the points: it's built into the digital SAT for the whole Math section.
4. Answer in the shape asked, then check it against every condition in the question.

Each section below turns these moves into the steps for its question type. The Desmos course teaches the calculator itself.

If function questions are new to you, start with lesson 1.

Function notation questions give you one side of $f(a)=b$ and ask for the other. Given an input, like $f(-2)$, you evaluate. Given an output, like $f(a)=18$, you work backward, and a nonlinear function can send two or more inputs to the same output. Before any algebra, ask which way the question goes.

How to answer function notation questions

1. Read which way it goes: a known input means evaluate; a known output means solve.
2. Evaluate: put the whole input, in parentheses, everywhere $x$ appears, then follow the order of operations. When you need the same rule again and again, or one function inside another, define it once in Desmos and call it by name: see compose functions with Desmos.
3. Solve every branch: set the rule equal to the output and solve. $(x-1)^2=9$ splits into $x-1=3$ and $x-1=-3$. When a line $y=k$ might cross the graph more than once, graph both in Desmos first to count the solutions: see count equation solutions with Desmos.
4. Keep what the question allows: drop any input that makes a denominator zero, puts a negative under a square root, or breaks a condition like $a<3$.

Evaluating and working backward

For a form like $f(x)=ab^x+c$, subtract each output from the next: $c$ cancels, and dividing the second change by the first gives $b$. Outputs $7$, $11$, $23$ change by $4$, then $12$, so $b=\frac{12}{4}=3$, and then $a$ and $c$ follow as exact numbers. For one function inside another, like $p(p(x))=0$, define $p$ in Desmos and enter the whole equation: it shows every root at once, so no branch goes missing. See evaluate nonlinear functions SAT.

Mistakes to avoid with function notation

$(a-3)^2=16$ gives $a-3=4$ or $a-3=-4$, so $a=7$ or $a=-1$: write both before any condition rules one out. A negative input goes in parentheses: without them, $2(-2)^2$ became $2-2^2$ and $-3(-2)$ became $-3-2$. With them, $p(-2)=2(-2)^2-3(-2)+1=15$. And check every answer in the original: $\sqrt4$ is $2$, not $-2$, so squaring added an answer that doesn't work.

Graph questions show a graph, a table or an equation and ask for one feature: a zero, an intercept, a maximum or minimum, where the function is positive or increasing, or the line its graph approaches. You don't need to name the function first. Words first, then the graph: the wording tells you which feature, and which coordinate of it.

How to answer graph questions

1. Name the feature and translate it: a zero means $f(x)=0$, the $y$-intercept means $x=0$, a maximum is the greatest output, and solutions to $f(x)=k$ are where the line $y=k$ meets the graph.
2. Look where it shows. If a graph, a table or one quick step like $f(0)$ shows it, read it. If only the equation hides it, graph it in Desmos, click the curve, then click the gray points you need. See Desmos points of interest SAT.
3. Answer in the shape asked: an $x$-value, an output, a whole point or an interval.

Reading nonlinear graphs

The point $(a,b)$ on a graph says $f(a)=b$, the same fact as a table row. Positive means above the $x$-axis; increasing means climbing from left to right, and a curve can do one without the other. Two inputs with the same output sit the same distance from a parabola's axis of symmetry, so the axis is at their midpoint. An exponential like $2^x+1$ levels off toward its horizontal asymptote, $y=1$, and a rational function like $\frac{4}{x-2}+1$ has a vertical asymptote where its denominator is $0$. See SAT nonlinear graph features.

Mistakes to avoid with graph features

The question asked for a value of $x$, so the answer is $5$. Just right of the vertex, the curve climbs but is still below the axis: increasing and negative at once. And $x=2$ is an input the function can't take; its real $x$-intercept is $-2$, where $\frac{4}{-4}+1=0$.

Quadratic function questions give a parabola as an equation, a graph or a list of facts and ask for one feature: the vertex, the zeros, the $y$-intercept, a minimum or maximum. Others give the facts and ask for the equation. Each of the three forms shows a different feature, so let the question pick the form.

How to answer quadratic function questions

1. Read what it asks for: the $y$-intercept, the zeros, the vertex, a minimum or maximum, or the equation itself.
2. Pick the form that shows it. Standard form $ax^2+bx+c$ shows the $y$-intercept, $(0,c)$, and puts the vertex at $x=-\frac{b}{2a}$. Factored form $a(x-r)(x-s)$ shows the zeros $r$ and $s$, with the vertex midway between them. Vertex form $a(x-h)^2+k$ shows the vertex $(h,k)$, so the maximum or minimum is $k$. When the choices are whole functions, graph the original in Desmos and try each choice until one lands on top of it: see equivalent expressions with Desmos.
3. Find $a$. Roots or a vertex fill in everything but $a$, so put in one more point that isn't a root or the vertex. With three unrelated points, run a quadratic regression in Desmos instead: see Desmos regression.
4. Check every condition before you choose: the roots, the vertex, every given point, and which way the parabola opens.

Vertex, factored and standard form

The sign inside a factor or a square is the opposite of the input it gives: $(x+3)^2$ puts the vertex at $x=-3$. To get vertex form from standard form, complete the square: factor $a$ out of the $x$ terms, add the square of half the $x$-coefficient inside, and subtract that amount times $a$ outside. From factored form it's quicker: the vertex is at the midpoint of the zeros, $h=\frac{r+s}{2}$, and $k=f(h)$. See vertex form to standard form.

The $4$ added inside the parentheses is multiplied by $3$, so it adds $12$ to the function, not $4$.

Build a quadratic from its facts

A root at $-2$ means the factor $x+2$, and a vertex at $(2,-5)$ means $a(x-2)^2-5$. Then a point that isn't built in finds $a$: one equation, one unknown. If you know the axis of symmetry, mirror a point across it to find its partner: with the axis at $x=3$, a root at $-1$ has a partner at $7$. See build quadratic function from conditions SAT.

Mistakes to avoid with quadratic functions

$x+2$ is $x-(-2)$, so the vertex is $(-2,7)$. The vertex makes the squared term $0$, leaving $-5=-5$, which says nothing about $a$: use another point. And the sign of $a$ decides: negative opens down, so $7$ is a maximum.

Quadratic model questions put a parabola in a story: a ball's height over time, a profit by price. They ask what the vertex, an intercept or a coefficient means, when something peaks or lands, or for a prediction. A point isn't an answer until it's a sentence, with each number in its own unit.

How to answer quadratic model questions

1. Say what each axis means, with units: the input (seconds, items, dollars) and the output (meters, profit).
2. Find the point. Vertex form shows the vertex. Standard form hides it, and the landing time too, so graph the model in Desmos and click the vertex and the intercepts: see Desmos points of interest SAT.
3. Maximum or minimum: a negative number in front of the square makes a hill, so the vertex is a maximum; a positive one makes a valley and a minimum.
4. Put it in a sentence with each number in its unit, and keep only inputs the story allows: no negative times, no negative numbers of items.

Quadratic models and their extrema

In $h(t)=-3(t-2)^2+48$, the vertex $(2,48)$ says the object peaks at $48$ meters, $2$ seconds after launch, and $h(0)=36$ is its starting height. The $-3$ isn't a rate: the object falls $3$ meters in the first second after the peak, then $9$, then $15$, because $-3$ multiplies the squared time from the peak. The model's roots are $-2$ and $6$, but only $0\le t\le6$ describes the flight. See quadratic models and extrema SAT.

Mistakes to avoid with quadratic models

The first number takes the input's unit and the second the output's: $48$ meters at $2$ seconds. The fall grows every second, so $-3$ can't be a steady rate. And $t=-2$ is two seconds before launch: the object lands at the positive root, $t=6$.

Exponential questions describe something that grows or shrinks by the same factor over equal intervals, in a story, a table or an equation. They ask for the model, what a number in it means, or the factor over a different interval. Start, factor, interval: find those three, and the model writes itself.

How to answer exponential questions

1. Same ratio or same difference? Over equal steps, the same difference means linear and the same ratio means exponential.
2. Find the start, the factor and the interval. Growth by $r$, as a decimal, gives the factor $1+r$, and decay gives $1-r$: the factor is what remains. So $a(1+r)^t$ grows and $a(1-r)^t$ decays.
3. Write $a\,b^{t/k}$: $a$ is the start, $b$ the factor, and $\frac{t}{k}$ counts the $k$-unit intervals. Check it: at $t=k$, the model is $a\,b$.
4. Change the interval with a power, not by multiplying the percent. The new factor is $b^{\text{new}/\text{old}}$, and Desmos works it out in one line without rounding: see Desmos exponential models SAT.

Exponential growth and decay

A colony of $320$ that grows $15\%$ every $4$ days is $320(1.15)^{t/4}$. In $m(t)=80(0.72)^{t/6}$, $72\%$ remains every $6$ units, so $28\%$ is lost. For the percent lost every $5$ hours when the $20$-hour factor is $0.4096$, type (1-0.4096^(5/20))*100 into Desmos: $20$. See exponential growth and decay SAT.

Mistakes to avoid with exponential models

The factor is what's left: $0.72$ keeps $72\%$ and loses $1-0.72=28\%$. The exponent counts $4$-day intervals, so it's $\frac{t}{4}$. And each loss comes out of a smaller amount: four $20\%$ losses leave $0.8^4\approx0.41$, about a $59\%$ loss.

Transformation questions move or reshape a graph you know: shift it, reflect it or stretch it. They describe the moves in words and ask for the new rule, give a rule like $g(x)=-f(x-3)+1$, or ask where one named point lands. Solve the inside, then apply the outside.

How to answer transformation questions

1. Sort inside from outside: a change inside $f(\ )$ changes the input, the $x$-coordinate; a change outside changes the output, the $y$-coordinate.
2. Write the rule as $g(x)=A f(x-h)+k$. $x-h$ shifts right $h$, and $x+h$ shifts left. A negative $A$ reflects across the $x$-axis, and $|A|$ stretches or compresses. $+k$ shifts up $k$.
3. For a point $(u,v)$, set the inside equal to $u$ and solve for the new $x$; the new output is $Av+k$.
4. Check a point. When you know the rule for $f$ and the question asks about the new graph, define both in Desmos and click the point you need: see compose functions with Desmos.

Transform nonlinear functions

For $g(x)=-2f(x-3)+1$, the point $(3,2)$ on $f$ lands at $x=6$, since $x-3=3$, and $y=-2(2)+1=-3$. A harder inside works the same way: with $g(x)=4-2f(7-3x)$ and $f(5)=-2$, solve $7-3x=5$ for $x=\frac23$, and the output is $4-2(-2)=8$. See SAT function transformations.

Mistakes to avoid with transformations

Inside, the new input has to solve $x-h=u$, so a shift right is $x-3$. Outside, multiply first, then add: $-2(2)+1=-3$. And a messy inside isn't read off: set it equal to the old input and solve.