Digital SAT Math · Advanced Math
Interpreting nonlinear graphs and models
Most nonlinear items on the Digital SAT never ask you to complete the square or write a growth factor. They hand you a curve — or a short description of one — and ask what a feature means: where the graph starts, how high it gets, when it peaks, over which inputs it is rising, or what happens as time runs long. The algebra lives on the sibling pages; here the work is reading. Label the axes, locate the intercepts and the turning point, translate the question word into a graph feature, and only then compute if the picture will not give the number directly.
On the test
| Domain | Advanced Math (score report) |
| What it looks like | A parabola or exponential curve in the coordinate plane, or a worded model whose graph is described by intercepts and a turning point |
| Often asked | “What is the maximum value…?”, “At what … does … reach…?”, “Over what interval is the function increasing?”, “How many times…?”, “Which statement best describes…?”, “What is the best interpretation of…?” |
| Format | Multiple choice and student-produced response; grid-ins often want one coordinate of a vertex while the other coordinate sits in the common-error pile |
| Calculator | Rarely the bottleneck — the answer is usually a labeled point or a half-line you can see. Desmos helps only when a printed model is ugly and the item wants a number you cannot read |
Recognition cues: a curve that turns once (quadratic) or bends without turning (exponential); axis labels with units (seconds, dollars, grams); the words maximum, minimum, intercept, increasing, decreasing, approaches, how many times; and any stem that describes a graph without printing an equation.
What this page owns
The College Board files quadratic and exponential behaviour under one “nonlinear functions” heading. We already split the algebra of those curves onto quadratic functions (three forms, vertex formula, rebuild from intercepts) and exponential functions (percent to multiplier, periods in the exponent). This page owns the picture-first half of the same domain: you are given the graph (or a description of its features) and asked to name, locate, or interpret a feature without rewriting the rule. When the answer is a root of an equation you solve, that is nonlinear equations. When the graph is a straight line, that is linear functions. Form-to-feature algebra — “which equivalent form displays the vertex” — stays on the quadratic page; here the form is the graph itself.
Pattern recognition
Six shapes cover nearly every item:
- Read a marked point — intercept, vertex, or labeled height is the answer as printed.
- Value versus location — the question wants either the max/min value (a ) or where it occurs (an ); both numbers are always nearby.
- Interval of increase or decrease — the side of the vertex where the curve is rising or falling.
- Count crossings — how many times the graph reaches a height, meets a horizontal line, or hits zero.
- End behaviour — what an exponential does as the input grows without bound (approaches a floor; never turns around).
- Interpret in context — the same feature, restated in the units on the axes (empty tank, peak revenue, launch height).
Method
- Label the axes in the story’s units. Seconds on the bottom and feet on the side make “when” and “how high” impossible to swap. If the stem is pure and , still name which axis is the input.
- Locate the intercepts and the turning point, and say what each means. -intercept = value at input zero. -intercept = input where the output is zero. Vertex = highest or lowest point, and the sign of the opening tells you which.
- Translate the question word into a feature using the table above. Do this before you compute. Most wrong answers on this skill are a correct number attached to the wrong feature.
- Read first; compute only if the feature is not labeled. A marked vertex is a free answer. An unmarked height may need one substitution into a printed model, or a symmetry argument from two equal outputs.
- For intervals, stay on the input axis. Increasing and decreasing describe how outputs change as inputs move — the boundaries are -values (or -values), never the maximum height.
- For end behaviour, refuse the parabola instinct. An exponential that decays toward the axis does not turn around, does not go negative, and does not “reach” zero at a finite time.
| Feature on the graph | What it answers | Common misread |
|---|---|---|
| -intercept | starting / initial value | treated as a time or a max |
| -intercept | when the quantity is zero | treated as the starting value |
| Vertex | max or min value | reported when the stem asked where |
| Vertex | where the max or min occurs | reported when the stem asked what |
| Side of the vertex | interval of increase / decrease | flipped; or compared to the max |
| Horizontal line vs peak | how many solutions to | counted as one whenever a peak exists |
| Flattening toward an axis | end behaviour of a decay | read as a second turning point |
Worked example 1 — maximum value versus where it occurs
Stem. The graph of a quadratic function opens downward and has its vertex at . It crosses the -axis at . (a) What is the maximum value of ? (b) At what value of does reach that maximum?
Step 1 — name the feature each part wants. Part (a) asks for a value, so an output. Part (b) asks at what , so an input. Both answers live in the single ordered pair already given as the vertex.
Step 2 — read the vertex. The highest point is . The maximum value is therefore , and it occurs at .
Step 3 — refuse the other marked number. The -intercept is a real value of , but it is not the greatest — the curve still climbs from up to .
Check. Opening downward means the vertex is a max, not a min ✓. The two coordinates are different kinds of answer, so swapping them would reverse (a) and (b) ✓. Answers: (a) ; (b) .
Trap watch. Reporting for part (a) answers the location question. Reporting for part (a) confuses the intercept with the peak. Reporting for part (b) answers the value question. All three mistakes show up as lettered options on the practice set.
Worked example 2 — interval of increase from a described graph
Stem. The graph of a quadratic function opens downward and has its vertex at . Over which interval is increasing?
Step 1 — fix the geometry. Downward opening → the curve rises as it approaches the vertex from the left and falls as it leaves to the right.
Step 2 — write the interval on the input axis. Increasing means every input less than the vertex’s -coordinate:
(or, in interval notation, ). The number is the maximum value and does not appear in the answer.
Step 3 — sanity-check the other half. For the outputs fall, so that half-line is the interval of decrease. An interval that straddles mixes rising and falling and cannot be the answer to either question.
Check. Pick a test point left of the vertex, say : on a downward parabola with peak at , must be less than , and moving from toward raises the output ✓. Answer: .
Trap watch. is the decreasing half. or treats the maximum value as if it were on the -axis. “All real ” would require a line with positive slope, not a turning curve.
Worked example 3 — end behaviour and an intercept in context
Stem. The mass , in grams, of a radioactive sample hours after measurement begins follows a decreasing curve that starts at and passes through , , and , flattening toward the horizontal axis. (a) What is the best interpretation of the point ? (b) What happens to the mass as increases without bound?
Step 1 — read the -intercept in units. At the mass is grams, so is the initial mass when measurement begins — not a time of hours and not a long-run value.
Step 2 — read the pattern of outputs. Each hour the mass halves: . That is exponential decay toward zero, not a parabola about to turn around.
Step 3 — state the end behaviour. As grows without bound, approaches grams from above. It never becomes negative, and it never reaches a second peak.
Check. An initial value lives at input zero ✓. Halving each step stays positive forever ✓. No marked point is a turning point ✓. Answers: (a) the sample’s mass was grams at the start of measurement; (b) the mass approaches grams.
Trap watch. Calling a maximum that the sample “reaches and then exceeds” imports parabola language into an exponential. Reading the intercept as “empty after hours” confuses a -value with an -intercept. Predicting a negative mass later ignores that a positive quantity times a positive factor stays positive.
Practice
Answer before opening the explanation. Two items are student-produced response, matching the grid-in share of live nonlinear sets; six hand you a curve because on this skill the graph is the question, not an illustration. Every wrong choice below is one named misread — a value reported where a location was asked, an -intercept treated as a start, the wrong side of a vertex, or a height read off the axis instead of the curve. When you miss one, log the misread, not the item number.
Question 1 Warm-up
The graph of a quadratic function y = f(x) in the xy-plane opens downward, crosses the y-axis at (0, 5), and reaches a highest point at (3, 14). What is the starting value of f when x = 0?
Show the answer Choice A
Why it is right
The starting value when x = 0 is the y-intercept of the graph — the height of the curve where it meets the vertical axis. The stem states that the graph crosses the y-axis at (0, 5), so f(0) = 5. No algebra is required: the feature asked for is already named and given by the point in the stem.
Why each other choice fails
- Choice B
- Reports the maximum value of f, which is the y-coordinate of the highest point (3, 14). That answers a different question — how high the graph gets — not the value at x = 0.
- Choice C
- Reports the x-coordinate of the vertex, which is where the maximum occurs rather than any output of the function. The number 3 is an input, not a starting value.
- Choice D
- Guesses that a starting value must be zero. The y-intercept is given as (0, 5); the output at the start of the domain is 5, not 0.
Question 2 Standard
The graph of y = f(x) is shown in the xy-plane, where f is a quadratic function. What is the maximum value of f?
Show the answer Choice B
Why it is right
The parabola opens downward, so its highest point is the marked vertex (3, 16). The maximum value of a function is an output, so it is the y-coordinate of that highest point: 16. The picture confirms the reading — the curve is symmetric about the vertical line through x = 3, and no marked or visible point sits above y = 16.
Why each other choice fails
- Choice A
- Reports the x-coordinate of the vertex, which is where the maximum occurs rather than what it is. Substituting x = 3 into the graph gives the output 16, which is what the question asked for.
- Choice C
- Reads the y-intercept (0, 7) as the highest point. The graph continues to climb to the right of the y-axis until x = 3, so 7 is a value the function takes on the way up, not its greatest.
- Choice D
- Takes an x-intercept's x-coordinate. The curve crosses the x-axis at (-1, 0); -1 is neither the location nor the value of the maximum.
Question 3 Standard
The graph of a quadratic function f opens downward and has its vertex at (3, 16). Over which of the following intervals is f increasing?
Show the answer Choice C
Why it is right
A downward-opening parabola rises as you approach the vertex from the left and falls as you leave the vertex to the right. So f is increasing on the open half-line to the left of the vertex: every input less than 3. The y-coordinate of the vertex (16) has nothing to do with the interval of increase — that interval is measured on the x-axis.
Why each other choice fails
- Choice A
- This is the interval of decrease. To the right of the vertex the outputs fall, so f is decreasing for x > 3, not increasing.
- Choice B
- Uses the maximum value 16 as if it were a boundary on the x-axis. Intervals of increase are sets of inputs; 16 is an output, and comparing x to 16 does not describe where the curve is rising.
- Choice D
- A downward parabola is not increasing everywhere. It rises on one side of the vertex and falls on the other; only a strictly linear function with positive slope would increase on all real numbers.
Question 4 Standard
The graph shows the height h, in feet, of a model rocket t seconds after it is launched. At what value of t does the rocket reach its maximum height?
Show the answer Choice D
Why it is right
The question asks when the maximum occurs — an input measured in seconds — not how high the rocket gets. The marked peak of the graph is the point (4, 25), so the rocket reaches its maximum height at t = 4 seconds. The height itself is the other coordinate of that same point and answers a different question.
Why each other choice fails
- Choice A
- Reports the maximum height in feet rather than the time at which it occurs. The vertex is (4, 25); 25 is the value of h, not the value of t.
- Choice B
- Reads the positive t-intercept (9, 0), where the rocket returns to the ground. That is when height is zero, not when height is greatest.
- Choice C
- Reads the launch instant t = 0, where the height is 9 feet. The rocket is still climbing at launch; the peak is four seconds later.
Question 5 Standard
The graph shows the mass m, in grams, of a radioactive sample t hours after measurement begins. Which statement best describes the end behavior of the graph as t increases without bound?
Show the answer Choice A
Why it is right
The curve is a decaying exponential: it starts at the marked y-intercept (0, 80) and halves at each marked hour (40, 20, 10, …), flattening toward the horizontal axis without ever turning upward. As t grows without bound the outputs get arbitrarily close to 0 but stay positive — that is the end behavior of a positive decay model.
Why each other choice fails
- Choice B
- 80 is the initial mass at t = 0, not a long-run value. The graph leaves 80 immediately and never returns to it.
- Choice C
- Confuses this curve with a parabola. An exponential decay has no turning point and never begins to increase again; 80 is a starting height, not a maximum of a hill.
- Choice D
- A positive quantity multiplied by a positive factor stays positive. The graph approaches the axis from above and never crosses into negative mass.
Question 6 Harder
The graph of y = f(x) is shown in the xy-plane. For how many values of x does f(x) = 12?
Show the answer Choice B
Why it is right
The horizontal line y = 12 sits strictly below the peak (4, 20) and strictly above the marked points at height 4. A downward-opening parabola that rises above 12 and then falls back below it must cross y = 12 exactly twice — once on the way up and once on the way down. You do not need the exact x-values; the shape and the peak height are enough to count the intersections.
Why each other choice fails
- Choice A
- Would be correct only if the entire curve stayed below y = 12. The peak is at y = 20, well above 12, so the horizontal line must meet the graph.
- Choice C
- Would be correct if y = 12 were exactly the maximum value (a single tangency at the vertex). The maximum is 20, not 12, so the line cuts through the hill twice rather than touching once.
- Choice D
- A parabola is a degree-2 curve and meets a horizontal line in at most two points. Three intersections would require a higher-degree polynomial.
Question 7 Harder
The graph of the volume of water in a tank, in liters, as a function of time t in hours after noon, is a downward-opening parabola. The graph crosses the horizontal axis at t = 9 and has its highest point at (4, 50). Which statement is the best interpretation of the x-intercept at t = 9?
Show the answer Choice D
Why it is right
An x-intercept is a point where the output is zero. Here the output is volume in liters, so volume = 0 at t = 9 means the tank is empty 9 hours after noon. The starting volume would be the y-intercept (volume at t = 0), and the maximum volume is the vertex value 50 liters at t = 4 — neither of those is what an x-intercept reports.
Why each other choice fails
- Choice A
- Treats the x-intercept's number as a starting volume. A starting volume is a y-value at t = 0; the stem never gives the y-intercept, and 9 is an input (hours), not an output (liters).
- Choice B
- Confuses the x-intercept with the vertex. The maximum occurs at the marked highest point (4, 50), so the tank peaks at 4 hours after noon, not at 9.
- Choice C
- Invents a constant fill rate from the number 9. The graph is a parabola, so the rate of change is not constant, and 9 is a time when volume is zero rather than a slope.
Question 8 Harder Student-produced response
The graph of a quadratic function f opens upward, crosses the x-axis at (−2, 0) and (6, 0), and has its vertex at (2, −16). What is the minimum value of f?
Show the answer -16
Why it is right
The minimum value of an upward-opening parabola is the y-coordinate of the vertex. The stem places the vertex at (2, −16), so the minimum value is −16. The x-intercepts confirm the axis of symmetry is halfway between −2 and 6, which is x = 2, matching the given vertex; no further computation is needed once the vertex is known.
Answers students type instead
- 2
- Reports the x-coordinate of the vertex — where the minimum occurs — instead of the minimum value itself. The question asks for a y-value.
- 6
- Reports the other x-intercept. Same error as −2: an input where f is zero, not the minimum output.
- -2
- Reports one of the x-intercepts. An intercept is where the output is zero, not where the output is smallest.
Question 9 Harder
The graph of y = f(x) is shown in the xy-plane. Over which of the following intervals is f decreasing?
Show the answer Choice C
Why it is right
The marked vertex is the highest point (3, 16). To the right of x = 3 the curve falls: f(6) = 7 is already lower than f(3) = 16, matching the marked equal-height partner of (0, 7). A downward-opening parabola is decreasing on the entire half-line to the right of its vertex, so f is decreasing for all x > 3.
Why each other choice fails
- Choice A
- This interval straddles the vertex. On 0 < x < 3 the function is still increasing toward the peak; only the right half of the interval is decreasing. An interval of decrease cannot include a stretch where the outputs are rising.
- Choice B
- Compares inputs to the maximum value 16 as if 16 were on the x-axis. Intervals of increase or decrease are sets of x-values; the number 16 is an output, not a boundary for x.
- Choice D
- This is the interval of increase. Left of the vertex the curve is climbing toward (3, 16), so f is increasing for x < 3, not decreasing.
Question 10 Hardest
The graph of y = f(x) is shown in the xy-plane, where f is a quadratic function that opens downward and has vertex (3, 16). Which of the following is closest to f(1)?
Show the answer Choice B
Why it is right
At x = 1 the curve sits between the y-intercept (0, 7) and the vertex (3, 16). Because a parabola is symmetric about its axis, f(1) equals f(5). Reading the curve — or evaluating the rule that produces the marked points, f(x) = −x² + 6x + 7 — gives f(1) = −1 + 6 + 7 = 12. Among the choices, 12 is the exact value and is clearly closer to the curve at x = 1 than 16 (the peak), 7 (the intercept), or 0 (an intercept height).
Why each other choice fails
- Choice A
- Reads the maximum value off the vertex and treats it as the output at every nearby x. At x = 1 the graph has not yet reached the peak; the output is strictly less than 16.
- Choice C
- Reads the y-intercept (0, 7) and reports it as f(1). The intercept is the value at x = 0; one unit later the curve has already risen partway toward the vertex.
- Choice D
- Reads an x-intercept height. The outputs at the x-intercepts are 0, but those occur at x = −1 and x = 7, not at x = 1.
Question 11 Hardest Student-produced response
A model rocket is launched upward. Its height h, in feet above the ground, t seconds after launch is modeled by h(t) = −4t² + 24t + 5. At what time t, in seconds, does the rocket reach its maximum height?
Show the answer 3
Why it is right
The question asks when the maximum occurs — the t-coordinate of the vertex — not the maximum height itself. For h(t) = −4t² + 24t + 5 the vertex is at t = −b/(2a) = −24/(2 · −4) = −24/−8 = 3. Because a = −4 < 0 the parabola opens downward, so t = 3 is a maximum. (The height there is h(3) = −4(9) + 24(3) + 5 = 41 feet, which answers a different question.)
Answers students type instead
- 5
- Reports the launch height h(0) = 5. That is the starting height, not the time of the peak.
- 6
- Computes −b/a = 24/4 = 6, forgetting the 2 in the denominator of −b/(2a). That doubles the true vertex time.
- 41
- Computes the maximum height h(3) = 41 instead of the time at which it occurs. The question asks for t in seconds, not h in feet.
Question 12 Hardest
The graph shows the daily revenue R, in dollars, when a day-pass is priced at p dollars. Which statement is best supported by the graph?
Show the answer Choice A
Why it is right
The marked vertex of the revenue curve is (10, 500). Because the parabola opens downward, that point is a unique maximum: the greatest daily revenue is $500, and it occurs at a day-pass price of $10. Both coordinates matter, and the sentence must assign each to the correct unit — dollars of price on the horizontal axis, dollars of revenue on the vertical.
Why each other choice fails
- Choice B
- Swaps the coordinates of the vertex. A $500 day-pass price is far outside the model's zeros at p = 0 and p = 20, and $10 is not a revenue amount the graph ever marks as a peak.
- Choice C
- Treats the maximum revenue as a constant across the whole price range. At p = 5, for example, R(5) = −5(25) + 100(5) = 375, not 500; the peak is a single point, not a plateau.
- Choice D
- Treats revenue as if it rose by a fixed amount per dollar of price — a linear reading of a curved graph. Near the start the revenue gain per dollar is large and positive; after the peak it is negative. No $1 step increases revenue by $500.
Common mistakes
- Reporting the maximum value when the question asked where it occurs — the vertex is ; “at what time / price / ” wants , and is always in the choices.
- Reporting where it occurs when the question asked for the value — the same swap in the other direction; check units (seconds vs feet, dollars of price vs dollars of revenue).
- Treating an -intercept as the starting value — starting value is the -intercept (output at input zero). An -intercept is when the output is zero.
- Treating the -intercept as the maximum — only true if the vertex sits on the -axis. Otherwise the intercept is just another point on the way up or down.
- Judging “increasing” over the wrong interval — for a downward parabola the rising half is left of the vertex; flipping the inequality is the default trap.
- Writing an interval in terms of the max value — when the vertex is confuses an output with an input boundary.
- Counting one intersection whenever a peak exists — a horizontal line below the peak cuts a downward parabola twice; it cuts once only when the line is the peak height.
- Reading a -value off the axis instead of the curve — the tick mark nearest the point is not the output; follow the vertical line from the input up (or down) to the curve, then across.
- Expecting an exponential to turn around — end behaviour is a floor or a runaway, never a vertex. “Reaches a maximum and then increases again” is parabola language.
- Predicting a negative value for a positive decay model — mass, population, and price that decay toward zero stay non-negative.
- Swapping coordinates in a context sentence — “revenue of \10$500$” is the vertex pair with units reversed.
- Building a formula when the graph already answers — if the stem shows labeled points, the work is reading and translating, not completing the square.
FAQ
Do I need the equation of the curve? Only when the stem prints one and the feature you want is not labeled. A graph with a marked vertex is already the answer to both “what is the max” and “where does it occur.”
How do I know whether the vertex is a max or a min? Opening direction. Downward (the ends fall away from the turning point) means a maximum; upward means a minimum. Context helps too: a rocket’s height graph that returns to the ground is a downward parabola.
What is the difference between an -intercept and a root? None as points on the graph — both are inputs where the output is zero. The difference is framing: “root” is algebra language, “-intercept” is graph language. In context, that input is when the tank is empty, the rocket lands, or revenue hits zero.
Can a function be increasing on both sides of the vertex? Not if it has a single turning point. A parabola changes monotonicity exactly once. An exponential with base greater than 1 is increasing on its whole domain and has no vertex at all.
How do I count “how many times” without solving? Compare the target height to the peak. Above the peak: zero times. Equal to the peak: once. Strictly between the peak and the lower end of the window: twice for a single-turn parabola that crosses that height on both sides.
Grid-in format? Enter the exact value. The most common grid-in loss on this skill is entering the vertex’s other coordinate — there are no answer choices to warn you that you answered “where” when the item asked “what,” or the reverse.
How is this different from the quadratic-functions page? That page starts from a printed expression and rewrites it. This page starts from a picture (or a description of one) and names a feature. Same curves; different object of the question.