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🎯⭐ INTERACTIVE LESSON

Nonlinear Equations and Functions

Learn step-by-step with interactive practice!

Nonlinear Equations and Functions - Complete Interactive Lesson

Part 1: Function Notation

Functions & Graphs

Part 1 of 7 — Function Notation & Evaluation

What is a Function?

A function ff assigns exactly one output to each input. Written as f(x)=expressionf(x) = \text{expression}.

  • f(3)f(3) means "plug in x=3x = 3"
  • f(a+1)f(a + 1) means "replace every xx with (a+1)(a + 1)"

Example: If f(x)=2x2−3x+1f(x) = 2x^2 - 3x + 1:

f(4)=2(16)−12+1=21f(4) = 2(16) - 12 + 1 = 21

f(−1)=2(1)−3(−1)+1=2+3+1=6f(-1) = 2(1) - 3(-1) + 1 = 2 + 3 + 1 = 6

Domain & Range

  • Domain: all valid input values (check for division by zero, square roots of negatives)
  • Range: all possible output values

SAT Function Notation Tricks

f(x)=3f(x) = 3 asks: "For what value(s) of xx does the output equal 3?" This means solving f(x)=3f(x) = 3, NOT evaluating f(3)f(3).

On a graph: find where y=3y = 3 intersects the curve.


Worked Example 1

If f(x)=x2+3x−5f(x) = x^2 + 3x - 5, evaluate f(a+2)f(a + 2).

StepWork
Replace xx with (a+2)(a+2)f(a+2)=(a+2)2+3(a+2)−5f(a+2) = (a+2)^2 + 3(a+2) - 5
Expand (a+2)2(a+2)^2=a2+4a+4+3a+6−5= a^2 + 4a + 4 + 3a + 6 - 5
Simplify=a2+7a+5= a^2 + 7a + 5

Worked Example 2

The domain of f(x)=2x−6f(x) = \sqrt{2x - 6}.

StepWork
Expression under radical ≥0\geq 02x−6≥02x - 6 \geq 0
Solvex≥3x \geq 3
Domain[3,∞)[3, \infty)

Function Evaluation 🎯

Function Evaluation with Tables

The SAT often gives a table and asks you to evaluate:

xxf(x)f(x)
−1-144
0022
1100
22−2-2
3300

From this table: f(1)=0f(1) = 0, f(0)=2f(0) = 2, and f(f(1))=f(0)=2f(f(1)) = f(0) = 2.

Worked Example 3 — "Solving" f(x)=kf(x) = k

Using the table above, for what values of xx is f(x)=0f(x) = 0?

StepWork
Scan the f(x)f(x) column for 00f(1)=0f(1) = 0 and f(3)=0f(3) = 0
Answerx=1x = 1 and x=3x = 3

SAT Trap: If the question asks "f(x)=0f(x) = 0", do NOT evaluate f(0)=2f(0) = 2. Find where the output is 00.

Harder Function Notation 🎯

What Does the Notation Mean? 🔍

For each expression, choose what it represents.

Key Takeaways — Part 1

NotationMeaning
f(a)f(a)Substitute aa for every xx
f(x)=kf(x) = kSolve for xx — do NOT evaluate f(k)f(k)
f(0)f(0)The y-intercept
f(x)=0f(x) = 0The x-intercept(s)
DomainAll valid inputs (no ÷ 0, no negative\sqrt{\text{negative}})
RangeAll possible outputs
  • On graphs: f(a)=bf(a) = b means the point (a,b)(a, b) is on the curve
  • For expressions like f(a+1)f(a+1), replace every xx with (a+1)(a+1), then simplify

Part 2: Composition & Combining Functions

Functions & Graphs

Part 2 of 7 — Composition and Combining Functions

Composition: f(g(x))f(g(x))

"Evaluate inside out" — first compute g(x)g(x), then plug the result into ff.

Example: f(x)=x2f(x) = x^2 and g(x)=x+3g(x) = x + 3

f(g(2))=f(5)=25f(g(2)) = f(5) = 25

g(f(2))=g(4)=7g(f(2)) = g(4) = 7 — order matters!

Substituting an Expression: f(x+1)f(x + 1), f(2x)f(2x)

f(anything)f(\text{anything}) means "replace every xx in the rule with that anything, in parentheses."

  • If f(x)=3x−4f(x) = 3x - 4, then f(2x)=3(2x)−4=6x−4f(2x) = 3(2x) - 4 = 6x - 4
  • f(x+1)f(x + 1) is not the same as f(x)+1f(x) + 1: the first changes the input, the second changes the output

Combining Functions

NotationMeaning
f(x)+g(x)f(x) + g(x)Add the two outputs
f(x)−g(x)f(x) - g(x)Subtract the outputs (distribute the minus sign)
f(x)⋅g(x)f(x) \cdot g(x)Multiply the outputs

Worked Example 1

If f(x)=x2−3xf(x) = x^2 - 3x, find f(x+2)f(x + 2).

StepWork
Replace every xx with (x+2)(x + 2)(x+2)2−3(x+2)(x + 2)^2 - 3(x + 2)
Expand the squarex2+4x+4−3(x+2)x^2 + 4x + 4 - 3(x + 2)
Distribute the −3-3x2+4x+4−3x−6x^2 + 4x + 4 - 3x - 6
Resultf(x+2)=x2+x−2f(x + 2) = x^2 + x - 2

Worked Example 2

If f(x)=x2+1f(x) = x^2 + 1 and g(x)=3x−2g(x) = 3x - 2, find f(g(x))f(g(x)).

StepWork
Start with g(x)g(x)g(x)=3x−2g(x) = 3x - 2
Plug into fff(3x−2)=(3x−2)2+1f(3x - 2) = (3x-2)^2 + 1
Expand=9x2−12x+4+1= 9x^2 - 12x + 4 + 1
Simplify=9x2−12x+5= 9x^2 - 12x + 5

Composition & Combining Functions 🎯

Composition with Tables

The SAT frequently gives two tables and asks for a composition:

xxf(x)f(x)
13
25
31
xxg(x)g(x)
12
23
31

Find f(g(2))f(g(2)): g(2)=3g(2) = 3, then f(3)=1f(3) = 1. Answer: 11.

Find g(f(1))g(f(1)): f(1)=3f(1) = 3, then g(3)=1g(3) = 1. Answer: 11.

Working a Composition Backward

Sometimes the SAT gives the value of a composition and asks for the input.

Example: f(x)=2x+3f(x) = 2x + 3 and g(x)=x2g(x) = x^2. If f(g(k))=21f(g(k)) = 21 and k>0k > 0, find kk.

StepWork
Peel off the outer function2⋅g(k)+3=212 \cdot g(k) + 3 = 21, so g(k)=9g(k) = 9
Solve the inner functionk2=9k^2 = 9, so k=3k = 3 or k=−3k = -3
Apply the condition k>0k > 0k=3k = 3

Harder Composition 🎯

Composition Order Matters! 🔍

Given f(x)=x+1f(x) = x + 1 and g(x)=2xg(x) = 2x, evaluate each.

Key Takeaways — Part 2

ConceptKey Rule
f(g(x))f(g(x))Evaluate inside out — order matters!
f(x+1)f(x + 1)Replace every xx with (x+1)(x + 1)
f(x+1)f(x + 1) vs. f(x)+1f(x) + 1Changes the input vs. changes the output
f(x)⋅g(x)f(x) \cdot g(x), f(x)−g(x)f(x) - g(x)Combine the outputs; distribute any minus sign
Given f(g(k))f(g(k))Peel off the outer function first, then solve the inner one
TablesLook up values step by step

Part 3: Domain & Range

Functions & Graphs

Part 3 of 7 — Transformations of Functions

Vertical Transformations (Outside the function)

TransformationEquationEffect
Shift up kkf(x)+kf(x) + kGraph moves up kk units
Shift down kkf(x)−kf(x) - kGraph moves down kk units
Stretch by aa (if a>1a > 1)af(x)af(x)Graph gets taller
Compress by aa (if 0<a<10 < a < 1)af(x)af(x)Graph gets shorter
Reflect over x-axis−f(x)-f(x)Flip upside down

Horizontal Transformations (Inside the function)

TransformationEquationEffect
Shift right hhf(x−h)f(x - h)Graph moves right
Shift left hhf(x+h)f(x + h)Graph moves left
Compress by bbf(bx)f(bx)Graph gets narrower (b>1b > 1)
Reflect over y-axisf(−x)f(-x)Flip left-right

Key Insight

Horizontal transformations are opposite to what you might expect:

  • f(x−3)f(x - 3) moves the graph right, not left
  • f(2x)f(2x) makes the graph narrower, not wider

Worked Example 1

The graph of y=f(x)y = f(x) passes through (2,5)(2, 5). Where does the point move under y=3f(x−4)+1y = 3f(x - 4) + 1?

TransformationEffect on (2,5)(2, 5)
f(x−4)f(x - 4): right 4(2+4,5)=(6,5)(2 + 4, 5) = (6, 5)
3f3f: stretch yy by 3(6,15)(6, 15)
+1+ 1: up 1(6,16)(6, 16)

The point moves to (6,16)(6, 16).

Transformations 🎯

Combining Multiple Transformations

Apply transformations in this order:

  1. Horizontal shifts and stretches (inside)
  2. Reflections
  3. Vertical stretches (outside)
  4. Vertical shifts (outside)

Worked Example 2

Describe the transformations from y=∣x∣y = |x| to y=−2∣x+3∣+7y = -2|x + 3| + 7.

PieceTransformation
x+3x + 3Shift left 3
−2-2 (coefficient)Reflect over x-axis, stretch by factor 2
+7+7Shift up 7
VertexMoves from (0,0)(0, 0) to (−3,7)(-3, 7)
OpensDownward (because of the negative)

Worked Example 3

If f(3)=10f(3) = 10, what point must be on y=f(x+5)−2y = f(x + 5) - 2?

StepWork
Original point(3,10)(3, 10)
f(x+5)f(x + 5): left 5xx-coordinate: 3−5=−23 - 5 = -2
−2- 2: down 2yy-coordinate: 10−2=810 - 2 = 8
New point(−2,8)(-2, 8)

Applied Transformations 🎯

Name That Transformation 🔍

Identify the transformation applied to y=f(x)y = f(x).

Key Takeaways — Part 3

ModificationLocationDirection
f(x)+kf(x) + kOutsideUp kk (as expected)
f(x−h)f(x - h)InsideRight hh (opposite!)
af(x)af(x)OutsideVertical stretch/compress
f(bx)f(bx)InsideHorizontal compress (opposite!)
−f(x)-f(x)OutsideReflect over x-axis
f(−x)f(-x)InsideReflect over y-axis
  • To track a point: apply horizontal changes to xx, then vertical changes to yy
  • Vertex transformations: (0,0)→(h,k)(0,0) \to (h, k) in af(x−h)+ka f(x - h) + k

Part 4: Transformations

Functions & Graphs

Part 4 of 7 — Piecewise & Absolute Value Functions

Piecewise Functions

A function defined by different rules for different parts of its domain:

f(x)={x+3if x<0x2if x≥0f(x) = \begin{cases} x + 3 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}

f(−2)=−2+3=1f(-2) = -2 + 3 = 1 (use first rule since −2<0-2 < 0)

f(3)=9f(3) = 9 (use second rule since 3≥03 \geq 0)

Absolute Value as Piecewise

∣x∣={xif x≥0−xif x<0|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Graphing y=a∣x−h∣+ky = a|x - h| + k

  • V-shaped graph with vertex at (h,k)(h, k)
  • Opens up if a>0a > 0, opens down if a<0a < 0
  • Slope of right branch is aa, left branch is −a-a

Worked Example 1

Evaluate f(x)={2x−1x≤3x2−4x>3f(x) = \begin{cases} 2x - 1 & x \leq 3 \\ x^2 - 4 & x > 3 \end{cases} at x=3x = 3 and x=5x = 5.

InputWhich rule?CalculationResult
x=3x = 33≤33 \leq 3 → first rule2(3)−12(3) - 155
x=5x = 55>35 > 3 → second rule52−45^2 - 42121

SAT Graph Reading

When the SAT shows a piecewise graph, read each segment separately. Check:

  • What's the y-value at specific x-values?
  • Are the endpoints open circles (excluded) or closed circles (included)?

Piecewise Functions 🎯

Solving Absolute Value Equations

∣ax+b∣=c|ax + b| = c

Split into two cases: ax+b=cax + b = c or ax+b=−cax + b = -c (only when c≥0c \geq 0).

Worked Example 2

Solve ∣2x−5∣=7|2x - 5| = 7.

CaseEquationSolution
Positive2x−5=72x - 5 = 7x=6x = 6
Negative2x−5=−72x - 5 = -7x=−1x = -1

Both solutions: x=6x = 6 and x=−1x = -1.

Worked Example 3

For what values of xx is ∣x−4∣≤3|x - 4| \leq 3?

StepWork
Remove absolute value−3≤x−4≤3-3 \leq x - 4 \leq 3
Add 4 to all parts1≤x≤71 \leq x \leq 7
In interval notation[1,7][1, 7]

SAT Tip: ∣x−a∣≤d|x - a| \leq d means "xx is within dd units of aa." So ∣x−4∣≤3|x - 4| \leq 3 means xx is within 3 of 4.

Absolute Value Equations 🎯

Piecewise or Absolute Value? 🔍

Classify each function type and identify key features.

Key Takeaways — Part 4

ConceptKey Rule
PiecewiseCheck which condition xx satisfies, use that rule
$y = ax - h
$A
$A
$A
$x - a
Open vs closed circlesOpen = excluded, closed = included

Part 5: Function Composition

Functions & Graphs

Part 5 of 7 — Graph Analysis & Interpretation

Increasing vs. Decreasing

  • Increasing: as xx moves right, yy goes up
  • Decreasing: as xx moves right, yy goes down
  • Constant: horizontal line segment

Maximum and Minimum Values

  • Absolute max/min: the highest/lowest y-value on the entire graph
  • Relative (local) max/min: higher/lower than nearby points

On the SAT, these appear as:

  • "Over which interval is ff increasing?"
  • "At what value of xx does ff attain its maximum?"
  • "What is the maximum value of ff?" (asking for the y-coordinate)

Rate of Change

Average rate of change from x=ax = a to x=bx = b:

Rate=f(b)−f(a)b−a\text{Rate} = \frac{f(b) - f(a)}{b - a}

This is just the slope of the secant line between two points.


Worked Example 1

The table shows values of ff. Find the average rate of change from x=1x = 1 to x=5x = 5.

xx1122334455
f(x)f(x)337799881111
StepWork
Formulaf(5)−f(1)5−1\frac{f(5) - f(1)}{5 - 1}
Substitute11−34\frac{11 - 3}{4}
Simplify=2= 2

Note: The function goes up and down between x=1x = 1 and x=5x = 5, but the average rate of change only looks at endpoints.

Intercepts

  • x-intercepts: where f(x)=0f(x) = 0 (solve or read from graph)
  • y-intercept: evaluate f(0)f(0) (or read where graph crosses y-axis)

Graph Analysis 🎯

Comparing Rates of Change

The SAT may ask you to compare rates of change over different intervals.

Worked Example 2

Using the table:

xx0011223344
f(x)f(x)11449916162525

Where is ff increasing fastest?

IntervalRate of Change
[0,1][0, 1](4−1)/1=3(4 - 1)/1 = 3
[1,2][1, 2](9−4)/1=5(9 - 4)/1 = 5
[2,3][2, 3](16−9)/1=7(16 - 9)/1 = 7
[3,4][3, 4](25−16)/1=9(25 - 16)/1 = 9

ff increases fastest on [3,4][3, 4] with rate =9= 9.

(This pattern makes sense — it's f(x)=(x+1)2f(x) = (x+1)^2, a parabola that curves upward faster and faster.)

Positive, Negative, and Zero

FeatureMeaning on Graph
f(x)>0f(x) > 0Graph is above the x-axis
f(x)<0f(x) < 0Graph is below the x-axis
f(x)=0f(x) = 0Graph touches/crosses the x-axis

Deeper Graph Analysis 🎯

Reading the Graph 🔍

For a function with f(0)=2f(0) = 2, f(2)=6f(2) = 6, f(4)=6f(4) = 6, f(6)=0f(6) = 0:

Key Takeaways — Part 5

ConceptFormula / Rule
Average rate of changef(b)−f(a)b−a\frac{f(b) - f(a)}{b - a} = slope of secant line
Increasingff goes up as xx increases
Decreasingff goes down as xx increases
f(x)>0f(x) > 0Graph above x-axis
f(x)=0f(x) = 0Graph on x-axis (x-intercept)
"Maximum value of ff"The y-coordinate, not the x-coordinate
  • Zero average rate ≠\neq constant function — it just means endpoints match
  • Compare rates across intervals to find where the function changes fastest

Part 6: Exponential Functions & Graphs

Functions & Graphs

Part 6 of 7 — Exponential Functions and Their Graphs

The Form f(x)=a⋅bxf(x) = a \cdot b^x

  • aa is the initial value: f(0)=a⋅b0=af(0) = a \cdot b^0 = a, so the yy-intercept is (0,a)(0, a)
  • bb is the growth factor: each time xx goes up by 1, the output is multiplied by bb
  • b>1b > 1 → growth; 0<b<10 < b < 1 → decay

Reading the Percent from bb

FactorMeaning
b=1+rb = 1 + rIncreases by rr (as a percent) each step: 1.061.06 → up 6%6\%
b=1−rb = 1 - rDecreases by rr each step: 0.850.85 → down 15%15\%

Linear vs. Exponential

  • Linear: the output adds the same amount each step (constant difference)
  • Exponential: the output is multiplied by the same factor each step (constant ratio)

Worked Example 1

A function ff has f(0)=5f(0) = 5, f(1)=15f(1) = 15, f(2)=45f(2) = 45, f(3)=135f(3) = 135. Write f(x)f(x).

StepWork
Check differences10,30,9010, 30, 90 → not constant, so not linear
Check ratios15/5=45/15=135/45=315/5 = 45/15 = 135/45 = 3 → constant ratio
Initial valuef(0)=5f(0) = 5
Resultf(x)=5(3)xf(x) = 5(3)^x

Worked Example 2

For f(x)=800(0.75)xf(x) = 800(0.75)^x, describe the function and find f(2)f(2).

StepWork
Initial value800800
Factor0.75=1−0.250.75 = 1 - 0.25 → decreases by 25%25\% per step
Evaluatef(2)=800(0.75)2=800(0.5625)=450f(2) = 800(0.75)^2 = 800(0.5625) = 450

Exponential Basics 🎯

Features of Exponential Graphs

For f(x)=a⋅bx+kf(x) = a \cdot b^x + k with a>0a > 0:

FeatureHow to find it
yy-interceptf(0)=a+kf(0) = a + k
Level it approachesThe graph gets closer and closer to y=ky = k but never reaches it
xx-interceptSet f(x)=0f(x) = 0 and solve; there is none if k≥0k \geq 0

Worked Example 3

Find the intercepts of the graph of y=3(2)x−12y = 3(2)^x - 12.

StepWork
yy-intercept3(1)−12=−93(1) - 12 = -9 → (0,−9)(0, -9)
xx-intercept3(2)x=123(2)^x = 12 → 2x=42^x = 4 → x=2x = 2 → (2,0)(2, 0)

Worked Example 4 — Other Time Units

A quantity starts at 40 and doubles every 3 years: Q(t)=40(2)t/3Q(t) = 40(2)^{t/3}. Find Q(9)Q(9).

99 years is 9/3=39/3 = 3 doubling periods: Q(9)=40(2)3=320Q(9) = 40(2)^3 = 320.

Key insight: In bt/nb^{t/n}, the output is multiplied by bb once every nn units of tt.

Comparing Growth

g(x)=100+20xg(x) = 100 + 20x (linear) and h(x)=100(1.2)xh(x) = 100(1.2)^x (exponential) both start at 100 and both equal 120 at x=1x = 1. At x=5x = 5: g(5)=200g(5) = 200 but h(5)≈248.8h(5) \approx 248.8. An increasing exponential eventually passes any linear function.

Exponential Graphs & Models 🎯

Read the Exponential 🔍

Choose the correct description for each function.

Key Takeaways — Part 6

ConceptRule
f(x)=a⋅bxf(x) = a \cdot b^xaa = initial value (yy-intercept); bb = factor per step
Growth vs. decayb>1b > 1 grows; 0<b<10 < b < 1 decays
Percent changeb=1+rb = 1 + r (up rr) or b=1−rb = 1 - r (down rr)
Spotting exponential dataConstant ratio between outputs (linear has constant difference)
a⋅bx+ka \cdot b^x + kGraph approaches y=ky = k; yy-intercept is a+ka + k
bt/nb^{t/n}Multiplies by bb once every nn units of tt
  • An increasing exponential function eventually exceeds any increasing linear function

Part 7: Review & Applications

Functions & Graphs

Part 7 of 7 — Review & SAT-Level Mixed Practice

Functions Cheat Sheet

ConceptKey Idea
f(a)f(a)Substitute aa into the function
f(x)=kf(x) = kSolve for xx (or find where y=ky = k on graph)
f(g(x))f(g(x))Evaluate inside out
f(x+1)f(x + 1)Replace every xx with (x+1)(x + 1)
DomainAll valid inputs
RangeAll possible outputs
Increasingff goes up as xx moves right
a⋅bxa \cdot b^xaa = initial value; b>1b > 1 growth, 0<b<10 < b < 1 decay

SAT Strategies for Function Questions

  1. Use the answer choices — if asked for a function and given formulas, test with a value
  2. Read graphs carefully — pay attention to open vs. closed circles
  3. Don't confuse f(a)=bf(a) = b with f(b)=af(b) = a — in the first, aa is the input; in the second, aa is the output
  4. For word problems: identify input vs. output

Worked Example 1

If f(x)=ax+bf(x) = ax + b, f(2)=7f(2) = 7, and f(5)=16f(5) = 16, find aa and bb.

StepWork
Set up system2a+b=72a + b = 7 and 5a+b=165a + b = 16
Subtract3a=93a = 9 → a=3a = 3
Find bb6+b=76 + b = 7 → b=1b = 1
Answerf(x)=3x+1f(x) = 3x + 1

Worked Example 2

Find the range of f(x)=−2(x−3)2+8f(x) = -2(x - 3)^2 + 8.

StepWork
Identify vertex(3,8)(3, 8)
DirectionOpens down (a=−2<0a = -2 < 0)
Maximum value88 (at x=3x = 3)
Range(−∞,8](-\infty, 8] or f(x)≤8f(x) \leq 8

Mixed Review 🎯

Hard SAT Function Patterns

Pattern 1: Nested Function Evaluation

f(x)=x2−1f(x) = x^2 - 1. What is f(f(2))f(f(2))?

f(2)=3f(2) = 3, then f(3)=8f(3) = 8. Answer: 88.

Pattern 2: Functions Defined by Conditions

"f(x)f(x) is a linear function where f(3)=10f(3) = 10 and the rate of change is −2-2."

→ Slope =−2= -2: f(x)=−2x+bf(x) = -2x + b. f(3)=−6+b=10f(3) = -6 + b = 10 → b=16b = 16. So f(x)=−2x+16f(x) = -2x + 16.

Worked Example 3

ff is a quadratic with vertex (1,−4)(1, -4) that passes through (3,0)(3, 0). Find f(x)f(x).

StepWork
Vertex formf(x)=a(x−1)2−4f(x) = a(x - 1)^2 - 4
Plug in (3,0)(3, 0)0=a(4)−40 = a(4) - 4 → a=1a = 1
Finalf(x)=(x−1)2−4=x2−2x−3f(x) = (x - 1)^2 - 4 = x^2 - 2x - 3
Verifyf(3)=9−6−3=0f(3) = 9 - 6 - 3 = 0 ✓

Worked Example 4

If f(x)=3x−5f(x) = 3x - 5, for what value of xx does f(2x)=f(x)+10f(2x) = f(x) + 10?

StepWork
Expand f(2x)f(2x)3(2x)−5=6x−53(2x) - 5 = 6x - 5
Expand f(x)+10f(x) + 10(3x−5)+10=3x+5(3x - 5) + 10 = 3x + 5
Set equal6x−5=3x+56x - 5 = 3x + 5
Solve3x=103x = 10 → x=10/3x = 10/3

SAT-Level Challenge 🎯

Quick-Fire Function Review 🔍

Match each situation with the correct answer.

Key Takeaways — Full Functions & Graphs Review

TopicOne-Liner
Notationf(a)f(a) = plug in; f(x)=kf(x) = k = solve
CompositionInside out; order matters
Substitutionf(x+1)f(x + 1) changes the input; f(x)+1f(x) + 1 changes the output
TransformationsOutside = vertical; inside = horizontal (opposite)
PiecewiseCheck which rule applies at each xx
Absolute valueV-shape; $
Rate of changef(b)−f(a)b−a\frac{f(b)-f(a)}{b-a} = secant slope
Exponentiala⋅bxa \cdot b^x: start at aa, multiply by bb each step
DomainNo ÷ 0, no negative\sqrt{\text{negative}}

Final tip: On the SAT, always check whether the question asks for an xx-value or a yy-value. "At what xx..." vs. "What is the value of ff..." are different questions!