Nonlinear Functions
Interpret quadratic and exponential functions, find vertices, model growth and decay, and understand graph transformations with worked SAT examples.
Nonlinear Functions
SAT Math | Advanced Math | Foundation to Advanced
Nonlinear functions describe relationships that do not change at a constant rate. On the Digital SAT, you must understand quadratic and exponential functions, interpret their equations and graphs, identify important features, and use nonlinear models to solve real-world problems.
You will also learn how changing an equation affects its graph and how polynomial, rational, and radical functions differ from linear functions.
1. Understanding Nonlinear Functions
A function assigns exactly one output to every input in its domain.
A nonlinear function is a function whose graph is not a straight line.
Consider the linear function:
f(x) = 3x + 2
Whenever x increases by 1, the output increases by exactly 3.
Now consider:
g(x) = x²
x | g(x) |
|---|---|
0 | 0 |
1 | 1 |
2 | 4 |
3 | 9 |
4 | 16 |
The changes in output are:
1, 3, 5, 7
These differences are not constant.
Therefore, the function is nonlinear.
Common nonlinear functions include:
Quadratic functions, such as f(x) = x² + 3x − 4.
Exponential functions, such as f(x) = 2ˣ.
Polynomial functions, such as f(x) = x³ − 2x.
Rational functions, such as f(x) = 1/x.
Radical functions, such as f(x) = √x.
Key idea: A nonlinear function does not have one constant slope across its entire domain.
On the SAT, focus particularly on quadratic and exponential functions, which are central to the Nonlinear Functions skill.
2. Quadratic Functions
A quadratic function can be written in standard form:
f(x) = ax² + bx + c
where a, b, and c are constants and a ≠ 0.
The graph of a quadratic function is called a parabola.
For example:
f(x) = x² − 4x + 3
This is quadratic because the highest power of x is 2.
Understanding the coefficient a
If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward.
An upward-opening parabola has a minimum value.
A downward-opening parabola has a maximum value.
Understanding the coefficient c
In standard form:
f(x) = ax² + bx + c
the constant c is the y-intercept.
This is because:
f(0) = c
Worked Example 1 — Identifying a quadratic
Difficulty: Easy
The function f is defined by:
f(x) = 2x² − 5x + 8
What is the y-intercept of its graph?
Step 1: Identify the constant term.
c = 8
Step 2: Substitute x = 0 to verify.
f(0) = 2(0)² − 5(0) + 8
f(0) = 8
Final answer: The y-intercept is (0, 8).
SAT tip: When a quadratic is in standard form, the y-intercept is immediately visible.
3. The Vertex of a Parabola
The vertex is the turning point of a parabola.
For an upward-opening parabola, the vertex is its lowest point.
For a downward-opening parabola, the vertex is its highest point.
The vertex is written as an ordered pair:
(h, k)
where:
h is the x-coordinate.
k is the minimum or maximum output.
Vertex form
A quadratic function can be written as:
f(x) = a(x − h)² + k
Its vertex is:
(h, k)
Example
f(x) = (x − 3)² + 5
The vertex is (3, 5).
Since the coefficient of the squared term is positive, the parabola opens upward.
Therefore, the minimum value is 5.
Worked Example 2 — Finding the minimum
Difficulty: Easy
The function is:
f(x) = 2(x − 4)² + 7
What is the minimum value of f?
Step 1: Recognize vertex form.
f(x) = a(x − h)² + k
Step 2: Identify k.
k = 7
Step 3: Check the direction.
a = 2, which is positive.
The parabola opens upward.
Final answer: The minimum value is 7.
The minimum occurs when x = 4.
Important: The minimum value is the y-coordinate of the vertex, not the x-coordinate.
4. Finding the Vertex from Standard Form
When a quadratic is written as:
f(x) = ax² + bx + c
the x-coordinate of the vertex is:
x = −b/(2a)
Substitute this x-value into the function to find the y-coordinate.
Worked Example 3 — Finding the vertex
Difficulty: Medium
Find the vertex of:
f(x) = x² − 6x + 11
Step 1: Identify the coefficients.
a = 1
b = −6
c = 11
Step 2: Use the vertex formula.
x = −(−6)/(2 × 1)
x = 3
Step 3: Find f(3).
f(3) = 3² − 6(3) + 11
f(3) = 9 − 18 + 11
f(3) = 2
Final answer: Vertex (3, 2).
Because a > 0, the minimum value is 2.
Axis of symmetry
A parabola is symmetric around a vertical line passing through its vertex.
This line is called the axis of symmetry.
Its equation is:
x = −b/(2a)
For the previous example, the axis of symmetry is x = 3.
5. The Three Important Forms of a Quadratic
Quadratic functions can be written in different but equivalent forms.
Each form reveals different information.
Standard form
f(x) = ax² + bx + c
Most useful for identifying:
The coefficients.
The y-intercept.
Whether the parabola opens upward or downward.
Factored form
f(x) = a(x − r₁)(x − r₂)
Most useful for identifying:
The zeros or roots.
The x-intercepts.
Values of x where the output equals zero.
Vertex form
f(x) = a(x − h)² + k
Most useful for identifying:
The vertex.
The axis of symmetry.
The maximum or minimum value.
Worked Example 4 — Understanding factored form
Difficulty: Medium
The function is:
f(x) = (x − 2)(x − 7)
What are its x-intercepts?
Step 1: Set f(x) = 0.
(x − 2)(x − 7) = 0
Step 2: Apply the zero-product property.
x − 2 = 0
or
x − 7 = 0
Step 3: Solve.
x = 2 or x = 7
Final answer: (2, 0) and (7, 0).
The axis of symmetry lies halfway between the roots:
x = (2 + 7)/2 = 4.5
SAT tip: Use the form that directly reveals the feature requested instead of performing unnecessary calculations.
6. Maximum and Minimum Values
A quadratic function has one vertex.
Its maximum or minimum value depends on whether the parabola opens upward or downward.
Upward-opening parabola
Example:
f(x) = (x + 2)² + 3
Since the squared term is never negative:
(x + 2)² ≥ 0
Therefore:
f(x) ≥ 3
The minimum value is 3.
Downward-opening parabola
Example:
g(x) = −2(x − 5)² + 18
Since the squared term is never negative:
−2(x − 5)² ≤ 0
Therefore:
g(x) ≤ 18
The maximum value is 18.
Worked Example 5 — Maximum value
Difficulty: Medium
A quadratic function is defined by:
h(x) = −3(x + 2)² + 27
What is the maximum value of h?
Step 1: Identify the coefficient.
a = −3
The parabola opens downward.
Step 2: Identify the vertex.
h = −2
k = 27
Step 3: Determine the maximum.
The vertex gives the maximum output.
Final answer: 27.
This maximum occurs when x = −2.
7. Transformations of Quadratic Graphs
A transformation changes the position or shape of a graph.
The parent quadratic function is:
f(x) = x²
Vertical shifts
f(x) = x² + 4
The graph shifts upward by 4 units.
f(x) = x² − 4
The graph shifts downward by 4 units.
Horizontal shifts
f(x) = (x − 3)²
The graph shifts right by 3 units.
f(x) = (x + 3)²
The graph shifts left by 3 units.
Reflection
f(x) = −x²
The graph is reflected across the x-axis.
It opens downward instead of upward.
Vertical stretching
f(x) = 3x²
The graph becomes narrower than y = x² because its outputs are multiplied by 3.
Vertical compression
f(x) = 0.5x²
The graph becomes wider than y = x².
Worked Example 6 — Graph transformation
Difficulty: Medium
The graph of:
g(x) = (x − 5)² − 3
is obtained from the graph of f(x) = x² by which transformations?
Step 1: Identify the horizontal shift.
(x − 5) shifts the graph 5 units to the right.
Step 2: Identify the vertical shift.
−3 shifts the graph 3 units downward.
Final answer: Right 5 units and down 3 units.
The new vertex is (5, −3).
Common mistake: The sign inside the parentheses has the opposite direction from the horizontal shift.
8. Function Notation and Input-Output Values
Nonlinear functions use the same function notation as linear functions.
For example:
f(x) = x² + 2x − 3
To find f(4), substitute x = 4.
f(4) = 4² + 2(4) − 3
f(4) = 16 + 8 − 3
f(4) = 21
Therefore:
f(4) = 21
This means the point (4, 21) lies on the graph.
Finding the input
Sometimes the SAT gives an output and asks for the corresponding input.
Worked Example 7 — Finding inputs
Difficulty: Medium
For the function:
f(x) = x² − 4
what values of x satisfy f(x) = 12?
Step 1: Set the function equal to 12.
x² − 4 = 12
Step 2: Add 4.
x² = 16
Step 3: Take both square roots.
x = 4 or x = −4
Final answers: −4 and 4.
SAT tip: A quadratic function can have two different inputs that produce the same output.
9. Exponential Functions
An exponential function contains the input variable in an exponent.
A common form is:
f(x) = a(bˣ)
where:
a is the initial value.
b is the growth or decay factor.
x is the number of time periods or other input units.
In typical positive exponential growth and decay models, a > 0 and b > 0.
Exponential growth
If b > 1, the function grows as x increases.
Example:
f(x) = 100(1.2ˣ)
The factor 1.2 represents a 20% increase for each 1-unit increase in x.
Exponential decay
If 0 < b < 1, the function decreases as x increases.
Example:
f(x) = 100(0.8ˣ)
The factor 0.8 represents a 20% decrease for each 1-unit increase in x.
Understanding the initial value
When x = 0:
f(0) = a(b⁰)
Since b⁰ = 1:
f(0) = a
Therefore, a is the initial output.
Worked Example 8 — Evaluating exponential growth
Difficulty: Easy
The function is:
P(t) = 200(1.1ᵗ)
What is P(2)?
Step 1: Substitute t = 2.
P(2) = 200(1.1²)
Step 2: Calculate the power.
1.1² = 1.21
Step 3: Multiply.
P(2) = 200(1.21)
P(2) = 242
Final answer: 242
10. Percent Growth and Decay
SAT questions often describe growth or decay using percentages rather than giving the exponential equation directly.
Exponential growth formula
A(t) = P(1 + r)ᵗ
where:
P is the initial amount.
r is the growth rate expressed as a decimal.
t is the number of periods.
For 12% growth:
r = 0.12
Growth factor = 1.12
Exponential decay formula
A(t) = P(1 − r)ᵗ
For 12% decay:
r = 0.12
Decay factor = 0.88
Worked Example 9 — Population growth
Difficulty: Medium
A town has a population of 5,000. The population increases by 4% each year.
Which function models the population P(t) after t years?
Step 1: Identify the initial population.
P = 5,000
Step 2: Convert the growth rate into a decimal.
4% = 0.04
Step 3: Find the growth factor.
1 + 0.04 = 1.04
Step 4: Write the function.
P(t) = 5000(1.04ᵗ)
Worked Example 10 — Depreciation
Difficulty: Medium
A computer is worth $1,600 and loses 25% of its value each year.
What is its value after 2 years?
Step 1: Determine the decay factor.
1 − 0.25 = 0.75
Step 2: Write the model.
V(t) = 1600(0.75ᵗ)
Step 3: Substitute t = 2.
V(2) = 1600(0.75²)
Step 4: Calculate.
V(2) = 1600(0.5625)
Final answer: $900
Common mistake: A 25% decrease means multiplying by 0.75, not 0.25.
11. Exponential Functions from Tables
An exponential function has a constant multiplicative change for equal increases in input.
Consider:
x | f(x) |
|---|---|
0 | 3 |
1 | 6 |
2 | 12 |
3 | 24 |
Each output is twice the previous output.
Therefore, the growth factor is 2.
The initial value is 3.
The function is:
f(x) = 3(2ˣ)
Comparing linear and exponential tables
A linear function has constant differences.
An exponential function has constant ratios over equal input intervals.
For example:
Linear outputs: 4, 7, 10, 13
Differences: +3, +3, +3
Exponential outputs: 4, 8, 16, 32
Ratios: ×2, ×2, ×2
Worked Example 11 — Identifying an exponential model
Difficulty: Medium
An exponential function has the following values:
x | f(x) |
|---|---|
0 | 5 |
1 | 15 |
2 | 45 |
Which equation represents the function?
Step 1: Identify the initial value.
f(0) = 5
Therefore, a = 5.
Step 2: Find the growth factor.
15/5 = 3
45/15 = 3
Therefore, b = 3.
Step 3: Write the model.
f(x) = 5(3ˣ)
12. Doubling and Halving
Exponential models often involve repeated doubling or halving.
Doubling
If a quantity doubles every period:
A(t) = P(2ᵗ)
For example, if 50 bacteria double every hour:
A(t) = 50(2ᵗ)
After 3 hours:
A(3) = 50(8) = 400
Halving
If a quantity halves every period:
A(t) = P(0.5ᵗ)
For example:
A(t) = 640(0.5ᵗ)
After 3 periods:
A(3) = 640(0.125)
A(3) = 80
When doubling takes several periods
If a quantity doubles every 3 years, a suitable model is:
A(t) = P(2ᵗᐟ³)
Here t represents years.
The exponent t/3 counts the number of 3-year doubling intervals.
Worked Example 12 — Doubling interval
Difficulty: Hard
A population starts at 300 and doubles every 4 years.
What is its population after 12 years?
Step 1: Determine the number of doubling periods.
12/4 = 3
Step 2: Apply the doubling factor three times.
300 × 2³
Step 3: Calculate.
300 × 8 = 2400
Final answer: 2,400
13. Comparing Exponential Functions
Two exponential functions can have different initial values and growth rates.
For example:
f(x) = 100(1.1ˣ)
g(x) = 80(1.2ˣ)
Function f starts at 100.
Function g starts at 80.
However, g grows faster because its growth factor is 1.2 rather than 1.1.
Important idea
The function with the greater initial value does not necessarily have the greater output for every future input.
A function with a higher growth factor can eventually exceed another function that started larger.
Worked Example 13 — Comparing growth rates
Difficulty: Medium
Company A's revenue increases by 6% each year.
Company B's revenue increases by 9% each year.
Which company has the larger annual growth factor?
Company A:
1 + 0.06 = 1.06
Company B:
1 + 0.09 = 1.09
Since 1.09 > 1.06:
Final answer: Company B
This conclusion compares growth rates, not total revenues. Without knowing the initial revenues and elapsed time, you cannot necessarily determine which company earns more money.
14. Transformations of Exponential Functions
The graph of an exponential function can shift upward, downward, left, or right.
Consider the parent function:
f(x) = 2ˣ
Vertical shift
g(x) = 2ˣ + 3
The graph shifts upward by 3 units.
Horizontal shift
g(x) = 2^(x − 2)
The graph shifts right by 2 units.
Reflection
g(x) = −2ˣ
The outputs are multiplied by −1, reflecting the graph across the x-axis.
Horizontal asymptote
For:
f(x) = 2ˣ
the horizontal asymptote is:
y = 0
For:
g(x) = 2ˣ + 3
the horizontal asymptote is:
y = 3
An asymptote is a line the graph approaches in a limiting direction.
Worked Example 14 — Exponential shift
Difficulty: Hard
The function:
g(x) = 3ˣ − 5
is obtained by transforming f(x) = 3ˣ.
Describe the transformation and find the horizontal asymptote.
Step 1: Identify the vertical shift.
−5 means the graph moves downward 5 units.
Step 2: Determine the asymptote.
The original horizontal asymptote is y = 0.
After shifting down 5 units, it becomes:
y = −5
Final answer: Down 5 units; horizontal asymptote y = −5.
15. Other Nonlinear Function Graphs
Besides quadratic and exponential functions, the SAT may require you to interpret polynomial, rational, or radical function graphs.
Polynomial functions
Example:
f(x) = x³ − 4x
A cubic function can have turning points and multiple x-intercepts.
Factoring can help identify its zeros:
f(x) = x(x² − 4)
f(x) = x(x − 2)(x + 2)
Therefore, the x-intercepts are:
(−2, 0), (0, 0), and (2, 0)
Rational functions
Example:
f(x) = 1/(x − 3)
The function is undefined when x = 3.
The graph has a vertical asymptote at x = 3.
For this function, the horizontal asymptote is y = 0.
Radical functions
Example:
f(x) = √(x − 4)
The expression inside the square root must be nonnegative.
x − 4 ≥ 0
Therefore:
x ≥ 4
The graph begins at (4, 0) and extends rightward.
SAT tip: When interpreting nonlinear graphs, examine intercepts, turning points, shifts, and restrictions on the inputs.
16. Real-World Applications of Quadratic Functions
Quadratic functions are frequently used to model area, height, and other relationships with squared variables.
Worked Example 15 — Projectile height
Difficulty: Medium–Hard
The height of an object, in meters, is modeled by:
h(t) = −5t² + 20t + 2
where t is time in seconds.
What is the maximum height of the object?
Step 1: Identify the coefficients.
a = −5
b = 20
c = 2
Step 2: Find the time at the vertex.
t = −b/(2a)
t = −20/[2(−5)]
t = 2
Step 3: Calculate the height at t = 2.
h(2) = −5(2²) + 20(2) + 2
h(2) = −20 + 40 + 2
h(2) = 22
Final answer: 22 meters
The maximum occurs after 2 seconds.
Worked Example 16 — Maximizing area
Difficulty: Hard
A rectangle has a perimeter of 40 units.
Its width is x and its length is 20 − x.
What is the maximum possible area?
Step 1: Write the area function.
A(x) = x(20 − x)
Step 2: Expand.
A(x) = −x² + 20x
Step 3: Find the vertex.
x = −b/(2a)
x = −20/[2(−1)]
x = 10
Step 4: Calculate the area.
A(10) = 10(20 − 10)
A(10) = 100
Final answer: 100 square units
The maximum occurs when the rectangle is a 10-by-10 square.
17. Advanced SAT Strategies and Common Mistakes
Strategy 1 — Recognize the function type
If the input is squared, the relationship may be quadratic.
If the input appears in an exponent, the relationship may be exponential.
Do not confuse these two types.
Strategy 2 — Choose the useful form
Standard form reveals the y-intercept.
Factored form reveals the zeros.
Vertex form reveals the minimum or maximum.
Strategy 3 — Understand initial values
In f(x) = a(bˣ), the initial value is a.
In a quadratic standard form f(x) = ax² + bx + c, the initial output is c.
Strategy 4 — Convert percentages correctly
For 15% growth, use 1.15.
For 15% decay, use 0.85.
Strategy 5 — Distinguish additive and multiplicative changes
Linear functions change by a constant amount.
Exponential functions change by a constant factor over equal input intervals.
Strategy 6 — Read the question carefully
The SAT may ask for:
A function value.
A vertex coordinate.
A maximum or minimum output.
The initial value.
An exponential growth factor.
A percent growth or decay rate.
A graph transformation.
An intercept.
Do not answer a different quantity.
Strategy 7 — Use Desmos strategically
Desmos can help visualize parabolas, inspect intersections, and compare models.
However, identifying a value directly from vertex or factored form may be faster.
Common Mistake 1 — Confusing the vertex coordinates
For:
f(x) = (x − 3)² + 7
the vertex is (3, 7), not (−3, 7).
Common Mistake 2 — Confusing minimum value with where it occurs
For:
f(x) = 2(x − 4)² + 7
the minimum value is 7.
It occurs at x = 4.
Common Mistake 3 — Forgetting percentage conversion
A 20% increase means multiplying by 1.20.
A 20% decrease means multiplying by 0.80.
Common Mistake 4 — Treating exponential growth as linear growth
Increasing by 10% per year means multiplying by 1.10 each year.
It does not mean adding 10% of the original amount every year.
Common Mistake 5 — Reading the wrong coefficient
In:
f(x) = 3x² + 5x − 2
the y-intercept is −2.
The number 5 is not the y-intercept.
Common Mistake 6 — Assuming every quadratic has two x-intercepts
A quadratic graph can have zero, one, or two distinct x-intercepts.
Common Mistake 7 — Ignoring graph restrictions
A rational function may be undefined at certain inputs.
A square-root function may require its radicand to be nonnegative.
Common Mistake 8 — Misreading horizontal shifts
(x − 4)² shifts right by 4.
(x + 4)² shifts left by 4.
Common Mistake 9 — Comparing initial values instead of growth factors
A function can start smaller but grow more rapidly.
Common Mistake 10 — Solving unnecessarily
If a quadratic is already in vertex form, identify its vertex directly rather than expanding it.
18. Check Your Understanding
Question 1 — Quadratic vertex
What is the vertex of:
f(x) = (x − 5)² + 3?
A. (−5, 3)
B. (5, 3)
C. (3, 5)
D. (5, −3)
Correct answer: B — (5, 3)
The function is in vertex form, f(x) = a(x − h)² + k.
Therefore, h = 5 and k = 3.
Question 2 — Maximum value
What is the maximum value of:
g(x) = −2(x + 1)² + 12?
A. −2
B. −1
C. 12
D. 24
Correct answer: C — 12
The coefficient of the squared term is negative.
The parabola opens downward.
Its maximum value is the vertex's y-coordinate, 12.
Question 3 — Exponential growth
A quantity increases by 8% each year.
Which factor represents its annual growth?
A. 0.08
B. 0.92
C. 1.08
D. 1.80
Correct answer: C — 1.08
An 8% increase means multiplying by 1 + 0.08 = 1.08.
Question 4 — Exponential decay
A product worth $500 loses 20% of its value annually.
What is its value after one year?
A. $100
B. $300
C. $400
D. $480
Correct answer: C — $400
Decay factor = 1 − 0.20 = 0.80.
500 × 0.80 = 400.
Question 5 — Function table
An exponential function satisfies:
f(0) = 6
f(1) = 18
f(2) = 54
What is f(3)?
A. 60
B. 108
C. 162
D. 216
Correct answer: C — 162
The outputs are multiplied by 3 at each step.
54 × 3 = 162.
Question 6 — Quadratic minimum
What is the minimum value of:
f(x) = x² − 8x + 19?
A. 3
B. 4
C. 8
D. 19
Correct answer: A — 3
Find the vertex's x-coordinate:
x = −(−8)/(2 × 1) = 4
Evaluate:
f(4) = 16 − 32 + 19 = 3
19. Lesson Summary
A nonlinear function does not have a constant rate of change over its entire domain.
Quadratic functions have parabolic graphs.
The standard form of a quadratic is f(x) = ax² + bx + c.
Vertex form reveals the turning point and maximum or minimum value.
Factored form reveals the zeros or x-intercepts.
The vertex can be calculated using x = −b/(2a).
Quadratic graphs can shift, reflect, stretch, or compress.
Exponential functions are commonly written as f(x) = a(bˣ).
The number a represents the initial value, and b represents the multiplicative change factor.
Exponential growth occurs when b > 1, assuming a positive initial value.
Exponential decay occurs when 0 < b < 1, assuming a positive initial value.
Linear functions have constant additive changes, while exponential functions have constant multiplicative changes over equal input intervals.
Other nonlinear graphs may involve polynomial, rational, and radical expressions.
Always consider the graph's intercepts, turning points, transformations, and domain restrictions.
You are now ready to complete the Nonlinear Functions practice questions in Baloo Prep.