Graph and function foundations

Learning objectives

Move confidently among words, equations, tables, and pictures without changing the relationship.

By the end of this section

You should be able to read coordinates and scales, interpret function notation, distinguish inputs from outputs, identify intercepts and rates, and state which inputs and outputs a relationship allows. The objective is representation fluency, not memorizing pictures.

Coordinates, axes, and scales

An ordered pair (x, y) reports a horizontal coordinate first and a vertical coordinate second. Start at the origin (0, 0); move according to x, then according to y. Negative x is left, positive x is right; negative y is down, positive y is up.

Coordinate plane showing points (-3, 2) in quadrant II and (2, -2) in quadrant IV.xy−4−3−2−11234−3−2−1123(−3,2)(2,−2)

Read labels before shapes. Equal visual steps need not equal one unit. The horizontal axis might advance by 2 minutes per grid interval while the vertical axis advances by 50 liters. A graph’s visual steepness is not a numerical slope until the scales are accounted for.

A plotted point is one paired observation. If the horizontal axis is time in hours and the vertical axis is distance in kilometers, (3, 12) means 12 kilometers at 3 hours. It does not mean a speed of 12 kilometers per hour. A rate compares changes or, in a suitable proportional setting, distance to time.

A graph can show a relation without an equation. Read values supported by marked points, axis labels, and stated properties. A sketch or trend does not provide exact unmarked coordinates automatically. All diagrams in this guide use the stated scales; algebra and labeled data, not visual guesswork, establish exact answers.

Three checks before reading a graph

What does each axis measure? How much is one tick interval? Is the requested value an input, an output, an intercept, or a rate? Answer these before tracing any line.

Functions: one output for each allowed input

A function assigns exactly one output to each input in its domain. Different inputs may share an output. The relation (1, 4), (2, 4), (3, 5) can be a function. A relation containing both (1, 4) and (1, 7) cannot define y as a function of x, because one input would have two outputs. In function notation, f(3) means “the output of function f at input 3.” It is not f multiplied by 3. If f(x) = 2x + 1, then f(3) = 2(3) + 1 = 7. The letter inside the parentheses is a placeholder: the same rule could be written f(t) = 2t + 1.

Input 3 enters the rule 2x + 1 and produces output 7.Rule:2x+1Input:3Output:7

Two questions that look similar. “Find f(3)” supplies the input; substitute 3. “Find x such that f(x) = 3” supplies the output; find the input or inputs that produce it. For f(x) = 2x + 1, these answers are 7 and 1, respectively.

Tables give paired values. To read f(3), find input 3 and then its output. To solve f(x) = 3 from a table, search the output column for 3 and read the associated input. A table of selected values need not include every solution or every allowed input.

Graphs of functions. On y = f(x), a vertical line at a particular input meets the graph at at most one point. This is the vertical-line test. A horizontal line may meet a function graph more than once: multiple inputs can have the same output. A circle fails the vertical-line test as a full relation; a parabola opening upward passes it.

Symbol or phrase

Read it as

f (a) = b

Input a produces output b; the graph contains (a, b).

f(0)

The output at zero; if defined, it gives the graph’s vertical intercept.

f(x) = 0

Find inputs whose outputs are zero: the zeros or horizontal intercepts.

f(x) = g(x)

Find inputs where the two functions have equal outputs.

Connecting tables, equations, and graphs

A relationship may be described in words, listed in a table, encoded in an equation, or drawn as a graph. Each representation makes different information easy to see. A table reveals particular pairs; an equation computes outputs; a graph highlights intercepts, trend, and boundaries; a verbal description supplies meaning and units.

For a linear relationship, the rate of change is

m = change in outputchange in input = y2 − y1x2 − x1 (x2 ≠ x1).

Use the same subtraction order in the numerator and denominator. Dividing an output change by the number of rows is wrong when input intervals are unequal. A rate’s units are output units per input unit. In y = mx + b, m is the constant rate and b is the output when x = 0. A proportional relationship y = kx has constant ratio y/x = k for nonzero x and passes through the origin. A line with b ≠ 0 is linear but not proportional.

Representation

The same relationship

Words

A tank starts with 12 liters and gains 3 liters each minute.

Equation

V(t) = 12 + 3t, during the stated filling interval.

Selected pairs

(0, 12), (2, 18), (5, 27).

Graph

A straight line with vertical intercept 12 and rate 3 liters per minute.

What patterns do and do not prove

Equal output differences across equal input intervals are consistent with a line. Unequal slopes between listed pairs rule out one linear function through all those points. But a finite set of points lying on a line does not, by itself, prove that the unknown function is linear everywhere. A prompt stating “linear function” supplies that additional information.

A table with inputs 0, 1, 2 and outputs 2, 4, 8 suggests doubling, not a constant additive change. It is consistent with an exponential rule. Again, do not assert an unprovided global rule solely from a few data points.

Intercepts are locations, not just labels The vertical intercept is where x = 0. The horizontal intercept is where y = 0. For y = 2x − 6, the vertical intercept is (0, −6), and the horizontal intercept is (3, 0). The equation’s constant −6 is not automatically the horizontal intercept.

Domains, ranges, and boundaries

The domain is the set of allowed inputs. The range is the set of outputs actually produced by those inputs. Determine the domain before claiming a range. For f(x) = x2 on all real inputs, the range is y ≥ 0; restricting the inputs to 2 ≤ x ≤ 3 changes the range to 4 ≤ y ≤ 9.

Algebraic domains. A denominator cannot equal zero, and an even real root needs a nonnegative radicand. Thus 1/(x − 3) excludes x = 3, while x + 2 requires x ≥ −2. If both restrictions occur in one formula, satisfy both.

Contextual domains. A variable counting tickets may take only nonnegative integer values, possibly with a capacity limit. A variable measuring time may allow real numbers over a stated interval. Do not let an algebraically legal input override an explicit physical restriction.

Endpoint notation. A closed dot includes a boundary; an open dot excludes it. The interval −2 < x ≤ 4 can also be written (−2, 4]. Parentheses exclude an endpoint; square brackets include it. Infinity is not an endpoint value that can be included, so its interval notation always uses parentheses.

Number line shaded from an open endpoint at -2 to a closed endpoint at 4.−3−2−1012345−2<x≤4

Piecewise rules. A function can use different formulas on different input intervals. Choose the branch using the input condition, then calculate. At a boundary, inspect < versus ≤ so you use the correct branch. A piecewise description is still a function when every allowed input has exactly one output.

Read range vertically. Domain is the graph’s horizontal extent; range is its vertical extent. On an increasing segment with open lower-left endpoint and closed upper-right endpoint, both the smaller input and smaller output are excluded, while the larger ones are included. On a decreasing segment, left/right and lower/upper endpoints no longer correspond in that way; inspect each coordinate.

A complete function statement A rule, a domain, and a meaning belong together. “C(n) = 12 + 4n, where n is an integer from 0 to 6 and C is a cost in dollars” is more informative than the equation alone.

15 worked examples

4.01. Read a plotted point in the correct order

What are the coordinates of point A?

Point A is three units left of the origin and two units above it, at (-3, 2).xy−4−3−2−1123123A
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Recognize the structure. Read horizontal position first, vertical position second. Each grid interval represents one unit.

Work it through. Point A is three units left of the vertical axis and two units above the horizontal axis. Its horizontal coordinate is −3 and its vertical coordinate is 2.

Answer: (−3, 2).

Check. The point is left and above the origin, so its coordinate signs must be negative then positive.

Avoid the trap. (2, −3) reverses input and output and would lie in a different quadrant.

4.02. Read unequal axis scales without counting boxes as units

The plotted point gives the volume in a tank after 4 minutes. What is that volume?

Graph with horizontal time in minutes and vertical volume in liters; the plotted point is at 4 minutes and 150 liters.t(min)V(L)024650100150200
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Recognize the structure. One vertical grid interval represents 50 liters, not one liter.

Work it through. At t = 4, the point aligns with the vertical tick labeled 150.

Answer: 150 liters.

Check. The point is halfway between the 100- and 200-liter levels.

Avoid the trap. Counting three vertical boxes gives a position on the drawing, not a volume of 3 liters.

4.03. Evaluate function notation with a negative input

If f(x) = 2x2 − x + 1, what is f(−3)?

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Recognize the structure. f(−3) asks for the output at input −3. Replace every occurrence of x by the same complete value.

Work it through. Substitute with parentheses:

f(−3) = 2(−3)2 − (−3) + 1 = 18 + 3 + 1 = 22.

Answer: 22.

Check. The squared term contributes 18, and the remaining terms add 4. The output should therefore exceed 18.

Avoid the trap. Replacing only the first x, or interpreting −x as always negative, changes the rule. Here −(−3) is positive 3.

Key takeaway. Function evaluation is ordinary substitution with an input-output label. It is not multiplication by the function’s name.

4.04. Read a function value from a table

Selected values of f are shown. What is f(3)?

x

−2

0

3

5

f(x)

5

1

−5

−9

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Recognize the structure. The input is supplied inside the parentheses. Find 3 in the input row, then read the paired output.

Work it through. In the column with x = 3, the output is −5. No equation needs to be inferred from the other values.

Answer: −5.

Check. The table explicitly includes the pair (3, −5).

Avoid the trap. The output 5 occurs at input −2; finding a similar-looking number in the wrong row does not evaluate f(3).

Key takeaway. Use the information already supplied. Inferring a formula can add work and unjustified assumptions when the desired pair is listed directly.

4.05. Find an input when the output is specified

The function f is defined by f(x) = 2x − 3. For what value of x is f(x) = 11?

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Recognize the structure. The output is 11. This asks for an input, not for the value of f (11).

Work it through. Set the rule equal to the required output:

2x − 3 = 11 2x = 14 x = 7.

Answer: 7.

Check. f(7) = 2(7) − 3 = 11. Evaluating f (11) would instead produce 19, which answers a different question.

Avoid the trap. Do not substitute the requested output into the input slot. Translate “ f(x) = 11” into an equation before calculating.

Key takeaway. This input-output distinction is essential for zeros, inverse questions, intersections, and contextual models.

4.06. Distinguish a zero from the vertical intercept

For f(x) = 2x − 6, find the zero of f and the vertical intercept of y = f(x).

Line crossing the x-axis at 3 and the y-axis at -6.xy(3,0)(0,−6)
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Recognize the structure. A zero is an input producing output 0. The vertical intercept uses input 0.

Work it through. Solve 2x − 6 = 0 to get x = 3. Separately, evaluate f(0) = −6, giving the point (0, −6).

Answer: Zero: 3; vertical intercept: (0, −6).

Check. The two answers correspond to different marked points on the graph.

Avoid the trap. The constant term −6 is the output at zero, not the zero of the function.

4.07. Use changes in input, not the number of table rows

The table gives values of a linear function L. Find its rate of change and a formula for L(t).

t

1

4

10

L(t)

7

13

25

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Recognize the structure. The function is stated to be linear, but the input intervals are unequal.

Work it through. From t = 1 to t = 4, the rate is (13 − 7)/(4 − 1) = 6/3 = 2. From 4 to 10, it is (25 − 13)/(10 − 4) = 12/6 = 2. Write L(t) = 2t + b and use (1, 7): 7 = 2 + b, so b = 5.

Answer: Rate = 2; L(t) = 2t + 5.

Check. L(10) = 20 + 5 = 25, agreeing with the third pair.

Avoid the trap. The output differences 6 and 12 differ because the input intervals differ. That alone does not mean the function is nonlinear.

Key takeaway. A slope is a quotient of paired changes. If output is distance and input time, include distance-per-time units.

4.08. Use unequal slopes to rule out linearity

Can one linear function pass through all four pairs in this table?

x

0

1

2

3

y

1

3

7

13

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Recognize the structure. The input increases by 1 each time. A linear function would have the same output change for each such step.

Work it through. The successive output changes are 3 − 1 = 2, 7 − 3 = 4, and 13 − 7 = 6. Those unequal changes give unequal slopes, so no single line contains all four points.

Answer: No. The listed points are not consistent with one linear function.

Check. The line through the first two points is y = 2x + 1; at x = 2, it gives 5 rather than the listed 7.

Avoid the trap. A pattern can be regular without being linear. Constant second differences are not constant first differences.

4.09. Allow repeated outputs, but not conflicting outputs for one input

Relation A contains (−1, 4), (0, 4), (1, 5). Relation B contains (−1, 4), (−1, 5), (1, 6). Which defines y as a function of x?

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Recognize the structure. The restriction is one output per input. There is no requirement that every output have only one input.

Work it through. In A, the three distinct inputs each have one output. Inputs −1 and 0 both producing 4 is allowed. In B, input −1 is paired with both 4 and 5, so the output is not uniquely determined.

Answer: A is a function; B is not.

Check. A vertical line at x = −1 would meet B at two different heights. For A, every listed vertical line meets at most one point.

Avoid the trap. Rejecting A because an output repeats confuses the vertical-line test with a horizontal-line condition.

Key takeaway. This distinction explains why a quadratic function can produce the same output at two different inputs and still be a function.

4.10. Use context to restrict a formula’s input set

A workshop charges a $12 booking fee plus $4 per kit. A customer may order from zero through six kits. If C(n) = 12 + 4n gives the total cost, state the domain and range.

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Recognize the structure. n counts complete kits, so the allowed inputs are integers, not every real number in an interval.

Work it through. The domain is {0, 1, 2, 3, 4, 5, 6}. Applying the rule gives costs 12, 16, 20, 24, 28, 32, 36.

Answer: Domain: {0, 1, 2, 3, 4, 5, 6}; range: {12, 16, 20, 24, 28, 32, 36} dollars.

Check. The fee is charged even for the permitted zero-kit booking, giving C(0) = 12. Six kits cost 12 + 24 = 36.

Avoid the trap. Writing 12 ≤ C ≤ 36 alone includes costs such as $13 that this discrete model cannot produce.

Key takeaway. A formula’s algebraic domain can be larger than the domain allowed by its story.

4.11. Combine two different domain restrictions

For real outputs, what is the domain of g(x) = x+2x−1?

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Recognize the structure. The square root imposes one condition; the denominator imposes another. Both must hold.

Work it through. For a real numerator, require x + 2 ≥ 0, so x ≥ −2. For a defined quotient, require x − 1 ≠ 0, so x ≠ 1. Together these give x ≥ −2 with 1 excluded.

Answer: [−2, 1) ∪ (1, ∞), or all real x ≥ −2 except x = 1.

Check. At x = −2, the output is 0/(−3) = 0, so the left endpoint is allowed. At x = 1, the denominator is zero, so that input is not allowed.

Avoid the trap. Zero is allowed inside a square root but not as a denominator. Do not treat both restrictions as strict inequalities.

Key takeaway. Write restrictions before simplifying or solving; otherwise an invalid input can slip into the final result.

4.12. Read open and closed endpoints in two directions

The graph is a straight segment from an open point at (−2, 1) to a closed point at (4, 5). State its domain and range.

Increasing segment from an open point at (-2, 1) to a closed point at (4, 5).xy(−2,1)(4,5)
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Recognize the structure. Domain measures horizontal extent; range measures vertical extent. An open endpoint is excluded.

Work it through. Inputs run from −2 to 4, excluding −2 and including 4. Outputs run from 1 to 5, excluding 1 and including 5.

Answer: Domain: (−2, 4]; range: (1, 5].

Check. The included endpoint supplies both input 4 and output 5; the excluded endpoint supplies neither −2 nor 1 elsewhere on this increasing segment.

Avoid the trap. Do not copy the domain as the range. They refer to different coordinates.

4.13. Read equality of outputs as an intersection input

For which listed value of x is f(x) = g(x)?

x

0

1

2

3

f(x)

3

5

7

9

g(x)

9

8

7

6

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Recognize the structure. The two outputs must agree within the same input column.

Work it through. At x = 2, both functions have output 7. At the other listed inputs, the outputs differ. Thus the common graph point supplied by the table is (2, 7).

Answer: 2.

Check. f(2) = 7 = g(2). The response is the input 2, not the shared output 7.

Avoid the trap. The qualifier “listed” matters. Without additional rules, a table of selected values does not rule out other intersections at unlisted inputs.

Key takeaway. Solving a system graphically means finding a common point; the question may ask for its input, output, ordered pair, or another expression.

4.14. Choose the branch from the input condition

Shipping cost is modeled by C(w)={4,0 < w ≤ 2,4 + 1.5(w − 2),2 < w ≤ 10, where w is weight in kilograms and C is cost in dollars. Find C(2) and C(5).

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Recognize the structure. The input interval selects the formula. The boundary w = 2 belongs to the first branch.

Work it through. Because 0 < 2 ≤ 2, C(2) = 4. Because 2 < 5 ≤ 10, use the second rule: C(5) = 4 + 1.5(5 − 2) = 4 + 4.5 = 8.5.

Answer: C(2) = 4 and C(5) = 8.5 dollars.

Check. Above 2 kilograms, the model adds $1.50 for each extra kilogram. Five kilograms is 3 kilograms above the threshold.

Avoid the trap. Do not use 4 + 1.5(5); that would charge the extra rate on the first two kilograms as well.

Key takeaway. Interpret the condition first, then the formula. Boundary symbols determine which rule applies.

4.15. Know when two points do not determine an unknown function

A function f satisfies f(0) = 4 and f(2) = 10. No other information is given. What is f(1)?

A) 6 B) 7 C) 8 D) Cannot be determined

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Recognize the structure. Two points determine a line, but the problem does not say the function is linear.

Work it through. The rule f(x) = 3x +4 fits the given values and gives f(1) = 7. But the rule f(x) = x2 +x +4 also fits: it gives 4 at 0 and 10 at 2, while giving f(1) = 6. Two valid functions produce different answers.

Answer: D, cannot be determined from the information given.

Check. Both rules satisfy the two stated conditions. Their disagreement at input 1 proves that the desired value is not uniquely fixed.

Avoid the trap. Averaging the two outputs is justified here only with an additional linearity assumption. A neat pattern is not the same as supplied information.

Key takeaway. Distinguish a fact stated by the problem from a model you chose. Inference is valid only when its assumptions are warranted.

Section 4 readiness check

You should be able to translate f (a) = b into a point (a, b), read graph scales, find the correct coordinate at an intercept or intersection, and preserve domain restrictions. A representation is useful only when you understand what each axis, row, and symbol means.

Section-exit practice

Topic practice