Analyzing and Interpreting Graphs

Other examples of odd functions

• f(x) = x

• f(x) = x⁵

• f(x) = sin x

• f(x) = 1/x, for x ≠ 0

• f(x) = x³ − 4x

Worked Example

Determine whether:

f(x) = x³ − 4x

is odd.

Substitute −x:

f(−x) = (−x)³ − 4(−x)

f(−x) = −x³ + 4x

Factor out −1:

f(−x) = −(x³ − 4x)

Therefore:

f(−x) = −f(x)

The function is odd.

Functions That Are Neither Even Nor Odd

Some functions have neither y-axis symmetry nor origin symmetry.

Consider:

f(x) = x² + x

Substitute −x:

f(−x) = x² − x

This is not equal to f(x).

It is also not equal to −f(x).

Therefore, the function is neither even nor odd.

Important distinction

A graph can have a line of symmetry without being an even function.

For example:

f(x) = (x − 3)²

has a vertical line of symmetry:

x = 3

However, it is not symmetric about the y-axis, so it is not an even function.

Similarly, some graphs have rotational symmetry around a point other than the origin without being odd functions.

Symmetry comparison

The zero function f(x) = 0 is a special case: it is both even and odd.

Positive and Negative Intervals

Another important feature of a function is whether its outputs are positive or negative.

A function is positive wherever its graph lies above the x-axis.

A function is negative wherever its graph lies below the x-axis.

Mathematically:

• Positive: f(x) > 0

• Negative: f(x) < 0

• Zero: f(x) = 0

These intervals describe the sign of the function's outputs, not whether the graph is increasing or decreasing.

A function can be positive and decreasing at the same time.

Worked Example: A linear function

Consider:

f(x) = 2x − 6

First, find the x-intercept:

0 = 2x − 6

x = 3

The graph crosses the x-axis at (3, 0).

For x > 3:

2x − 6 > 0

Therefore, the function is positive.

For x < 3:

2x − 6 < 0

Therefore, the function is negative.

Positive interval: (3, ∞)

Negative interval: (−∞, 3)

At x = 3, the function equals zero, so this point belongs to neither interval.

Finding Positive and Negative Intervals of Quadratic Functions

Quadratic functions can have several sign patterns depending on their roots and the direction in which they open.

Worked Example: An upward-opening parabola

Consider:

f(x) = x² − 4

Factor:

f(x) = (x − 2)(x + 2)

The roots are:

x = −2 and x = 2

These divide the number line into three intervals:

• x < −2

• −2 < x < 2

• x > 2

Test one value in each interval.

Therefore:

Positive: (−∞, −2) and (2, ∞)

Negative: (−2, 2)

Worked Example: A downward-opening parabola

Consider:

g(x) = −x² + 9

Find the roots:

0 = −x² + 9

x² = 9

x = −3 or x = 3

Since the parabola opens downward, its graph lies above the x-axis between the roots and below the x-axis outside them.

Therefore:

Positive: (−3, 3)

Negative: (−∞, −3) and (3, ∞)

A parabola that never crosses the x-axis

Consider:

h(x) = x² + 2

Since x² ≥ 0:

x² + 2 ≥ 2

Therefore, the function is positive for every real x.

Positive: (−∞, ∞)

Negative: None

This demonstrates that a function does not need x-intercepts to have positive or negative intervals.

Sign Charts

A sign chart is a useful tool for identifying where a function is positive, negative, or zero.

Consider:

f(x) = (x + 3)(x − 1)

The roots are x = −3 and x = 1.

Multiplying two negative numbers gives a positive result.

Multiplying a positive number and a negative number gives a negative result.

Therefore:

Positive: (−∞, −3) and (1, ∞)

Negative: (−3, 1)

Sign charts are particularly useful for polynomial and rational functions.

For rational functions, remember to exclude any values where the denominator equals zero.

Distinguishing Sign from Direction of Change

One of the most important skills in graph analysis is distinguishing between the sign of a function and whether it is increasing or decreasing.

Consider:

f(x) = −x + 5

This function is decreasing everywhere because its gradient is −1.

However:

• It is positive for x < 5.

• It equals zero at x = 5.

• It is negative for x > 5.

The function is therefore positive and decreasing on (−∞, 5).

Similarly, the function:

g(x) = x − 5

is increasing everywhere but negative when x < 5.

These are different properties and should be analysed separately.

End Behaviour of Functions

End behaviour describes what happens to a function's outputs as x becomes very large and positive or very large and negative.

We examine two directions:

• As x approaches positive infinity.

• As x approaches negative infinity.

The symbol ∞ represents infinity.

The symbol −∞ represents negative infinity.

Infinity is not a number that a function eventually reaches. Instead, these expressions describe what happens as the input grows without bound.

End behaviour notation

We commonly write:

As x → ∞, f(x) → ∞

This means that as x becomes increasingly large and positive, the output increases without bound.

Similarly:

As x → −∞, f(x) → −∞

means that as x becomes increasingly large in the negative direction, the output decreases without bound.

End behaviour helps us understand the overall shape of a graph, especially outside the visible plotting window.

End Behaviour of Linear Functions

Consider:

f(x) = 2x + 3

The gradient is positive.

Therefore:

• As x → ∞, f(x) → ∞.

• As x → −∞, f(x) → −∞.

Now consider:

g(x) = −2x + 3

The gradient is negative.

Therefore:

• As x → ∞, g(x) → −∞.

• As x → −∞, g(x) → ∞.

The sign of the gradient determines the end behaviour of a nonconstant linear function.

End Behaviour of Quadratic Functions

For:

f(x) = ax² + bx + c

the sign of a determines the end behaviour.

Positive leading coefficient

If a > 0, the parabola opens upward.

Both ends rise:

As x → ∞, f(x) → ∞

As x → −∞, f(x) → ∞

Example:

f(x) = x² − 4x + 1

Negative leading coefficient

If a < 0, the parabola opens downward.

Both ends fall:

As x → ∞, f(x) → −∞

As x → −∞, f(x) → −∞

Example:

g(x) = −2x² + 3x + 5

The lower-degree terms affect the position and shape of the graph but do not change its end behaviour.

End Behaviour of Polynomial Functions

For polynomial functions, the highest-power term usually determines the end behaviour.

This term is called the leading term.

For example:

f(x) = 3x⁴ − 2x³ + 5x − 7

The leading term is:

3x⁴

Because the degree is even and the leading coefficient is positive, both ends rise.

Four common end behaviour patterns

Summary table

Worked Example

Describe the end behaviour of:

f(x) = −4x⁵ + 3x² − 8

The leading term is:

−4x⁵

The degree is odd.

The leading coefficient is negative.

Therefore:

As x → −∞, f(x) → ∞

As x → ∞, f(x) → −∞

The graph rises on the left and falls on the right.

End Behaviour of Exponential Functions

Exponential functions behave differently from polynomials.

Consider:

f(x) = 2ˣ

As x becomes increasingly large:

f(x) increases without bound.

Therefore:

As x → ∞, f(x) → ∞

As x becomes increasingly negative:

f(x) approaches zero from above.

Therefore:

As x → −∞, f(x) → 0

The line y = 0 is a horizontal asymptote.

Exponential decay

Consider:

g(x) = (1/2)ˣ

As x → ∞:

g(x) → 0

As x → −∞:

g(x) → ∞

The function decreases from left to right and approaches zero on the right.

Vertical translations and asymptotes

Consider:

h(x) = 2ˣ + 3

This is the graph of 2ˣ translated upward by 3 units.

Its horizontal asymptote is:

y = 3

Therefore:

As x → −∞, h(x) → 3.

As x → ∞, h(x) → ∞.

The vertical translation changes the horizontal asymptote but does not change the overall direction of exponential growth.

End Behaviour of Rational Functions

Rational functions are ratios of polynomial expressions.

Consider:

f(x) = 1/x

As x becomes very large and positive:

f(x) approaches zero from above.

As x becomes very large and negative:

f(x) approaches zero from below.

Therefore:

• As x → ∞, f(x) → 0.

• As x → −∞, f(x) → 0.

The graph has a horizontal asymptote:

y = 0

It also has a vertical asymptote at:

x = 0

The function is undefined at x = 0.

Worked Example

Consider:

g(x) = 3 + 2/x

As x → ∞:

2/x → 0

Therefore:

g(x) → 3

As x → −∞:

2/x → 0

Therefore:

g(x) → 3

The horizontal asymptote is:

y = 3

This illustrates how a function can approach a finite value at both ends rather than increasing or decreasing without bound.

End Behaviour of Trigonometric Functions

Periodic functions such as sine and cosine behave differently again.

Consider:

f(x) = sin x

The function repeatedly oscillates between −1 and 1.

As x approaches positive or negative infinity, the function continues oscillating.

It does not approach one particular value.

Therefore, sin x has no single limiting value as x → ∞ or x → −∞.

This demonstrates that not all functions have end behaviour described by approaching infinity or a horizontal asymptote.

Comparing Important Graph Features

When comparing functions, it is useful to organise the information into a table.

Consider these four functions:

A: f(x) = x²

B: g(x) = −x²

C: h(x) = x³

D: p(x) = 2ˣ

Comparison table

Comparing their end behaviour

f(x) = x²

Both ends rise.

g(x) = −x²

Both ends fall.

h(x) = x³

The left end falls and the right end rises.

p(x) = 2ˣ

The left end approaches zero, while the right end rises without bound.

Why comparisons matter

Although these functions may pass through some of the same points, their overall behaviour is very different.

For example, x² and x³ both pass through the origin, but:

• x² is even.

• x³ is odd.

• x² has a minimum.

• x³ has no local or absolute extrema over all real numbers.

• x² is never negative.

• x³ is negative for negative x-values.

Comparing several features gives a more complete understanding than looking at intercepts alone.

Summarising the Behaviour of a Graph

A complete graph description should combine several important features.

Rather than listing disconnected facts, we can explain how the function behaves as a whole.

Worked Example: A quadratic function

Consider:

f(x) = x² − 6x + 5

We will analyse its important features.

Finding the intercepts

For the x-intercepts, set f(x) = 0:

x² − 6x + 5 = 0

Factor:

(x − 1)(x − 5) = 0

Therefore:

x = 1 or x = 5

x-intercepts: (1, 0) and (5, 0)

For the y-intercept:

f(0) = 5

y-intercept: (0, 5)

Finding the turning point

Use:

x = −b/(2a)

Here a = 1 and b = −6.

x = 6/2 = 3

Substitute:

f(3) = 9 − 18 + 5

f(3) = −4

Vertex: (3, −4)

Since the parabola opens upward, this is a minimum.

Finding the symmetry

The axis of symmetry is:

x = 3

The graph is symmetric about this vertical line.

However, the function is neither even nor odd because it is not symmetric about the y-axis or origin.

Finding the positive and negative intervals

The roots are x = 1 and x = 5.

The parabola opens upward.

Therefore:

Positive: (−∞, 1) and (5, ∞)

Negative: (1, 5)

Finding increasing and decreasing intervals

The vertex occurs at x = 3.

Therefore:

Decreasing: (−∞, 3)

Increasing: (3, ∞)

Describing end behaviour

The leading term is x².

It has even degree and a positive leading coefficient.

Therefore:

• As x → −∞, f(x) → ∞.

• As x → ∞, f(x) → ∞.

Complete graph summary

This is a complete mathematical description of the function's main features.

Real-World Application: Projectile Motion

A ball is thrown upward from a platform.

Its height is modeled by:

h(t) = −5t² + 20t + 5

where:

• t is time in seconds.

• h is height in metres.

Initial height

At t = 0:

h(0) = 5

The ball begins 5 m above the ground.

Maximum height

The vertex occurs at:

t = −20/[2(−5)]

t = 2

Calculate:

h(2) = −5(2)² + 20(2) + 5

h(2) = −20 + 40 + 5

h(2) = 25

Maximum height = 25 m at t = 2 s

When the ball reaches the ground

Set h(t) = 0:

−5t² + 20t + 5 = 0

Divide by −5:

t² − 4t − 1 = 0

Using the quadratic formula:

t = 2 ± √5

The positive solution is:

t ≈ 4.24 s

Interpreting the function features

Over the physical flight:

• Height increases from t = 0 to t = 2 s.

• Maximum height is 25 m.

• Height decreases from t = 2 s to approximately t = 4.24 s.

• Height is positive before the ball lands.

• Height equals zero when the ball reaches the ground.

Although the mathematical quadratic has end behaviour extending to infinity, this is not physically meaningful for the ball's actual flight.

The practical domain is approximately:

0 ≤ t ≤ 4.24

This illustrates why real-world models must be interpreted within appropriate domains.

Real-World Application: Business Profit

Suppose a business models its profit using:

P(x) = −2(x − 30)² + 1800

where:

• x is the number of products sold.

• P(x) is profit in dollars.

Maximum profit

The vertex is:

(30, 1800)

Therefore:

Maximum modeled profit = $1,800

This occurs when 30 products are sold.

Positive and negative profit

A positive profit means the business earns more revenue than its costs.

A negative profit means the business operates at a loss.

Find the break-even points by setting P(x) = 0:

0 = −2(x − 30)² + 1800

2(x − 30)² = 1800

(x − 30)² = 900

x − 30 = ±30

x = 0 or x = 60

Therefore:

• Profit is positive for 0 < x < 60.

• Profit equals zero at x = 0 and x = 60.

• The mathematical model gives negative profit for x > 60.

Interpreting trends

Profit increases until 30 products are sold and decreases afterward.

The function is symmetric about:

x = 30

This means equal deviations above and below 30 produce equal modeled profits.

For example:

P(20) = −2(20 − 30)² + 1800 = 1600

P(40) = −2(40 − 30)² + 1800 = 1600

Both production quantities produce a modeled profit of $1,600.

The model is illustrative and assumes the quadratic relationship remains appropriate over the quantities being considered.

Real-World Application: Population Growth

Consider:

P(t) = 500(1.08)ᵗ

where t is time in years.

The population starts at 500 and increases by 8% each year.

Function features

• y-intercept: (0, 500).

• Increasing for all real t mathematically.

• Positive for all real t.

• No x-intercept.

• No maximum or minimum over all real t.

• Horizontal asymptote: P = 0.

• As t → ∞, P(t) → ∞.

• As t → −∞, P(t) → 0.

For a real population model, we might restrict the domain to t ≥ 0.

Over this practical domain, the minimum modeled population is 500 at t = 0.

This is another example of how restricting the domain changes the extrema of a function.

A Systematic Method for Analysing Any Graph

When given a graph or equation, use the following process.

Identify the function family.

Determine whether the graph is linear, quadratic, cubic, exponential, rational, trigonometric, or another type.

Find the domain and range.

Determine the allowed input values and possible outputs.

Find the intercepts.

Calculate or estimate where the graph meets the coordinate axes.

Investigate symmetry.

Check whether the graph has y-axis symmetry, origin symmetry, or another line of symmetry.

Identify positive and negative intervals.

Find where the graph lies above and below the x-axis.

Identify increasing and decreasing intervals.

Read the graph from left to right and locate changes in direction.

Locate extrema.

Identify local and absolute maximums and minimums where they exist.

Describe end behaviour.

Explain what happens as x approaches positive and negative infinity, when this is appropriate for the domain.

Identify asymptotes or discontinuities.

Check whether the graph approaches particular lines or contains gaps.

Summarise the overall behaviour.

Combine the important features into a clear mathematical description.

Common Mistakes

Confusing symmetry about the y-axis with symmetry about another vertical line

A parabola with axis x = 3 is not necessarily an even function.

Assuming every function is even or odd

Many functions are neither.

Confusing positive intervals with increasing intervals

Positive means the output is above zero. Increasing means the output rises as x increases.

Including roots in strictly positive or negative intervals

At an x-intercept, f(x) = 0, so the function is neither positive nor negative.

Describing end behaviour using only the visible graph

A plotting window may not show the long-term behaviour of a function.

Ignoring the leading coefficient

For polynomial functions, the sign of the leading coefficient affects the direction of the graph's ends.

Assuming all functions approach infinity at their ends

Exponential and rational functions may approach finite asymptotes, while trigonometric functions may oscillate indefinitely.

Confusing an asymptote with an intercept

An asymptote describes limiting behaviour, while an intercept is a point on the graph.

Assuming symmetry guarantees a maximum or minimum

For example, y = x³ has origin symmetry but no maximum or minimum over all real numbers.

Ignoring domain restrictions

A real-world model may only apply over a limited interval, changing which features are relevant.

Key Terms

• Function feature: A mathematical characteristic describing a graph or function.

• Symmetry: A property in which a graph matches itself under a reflection or rotation.

• Even function: A function satisfying f(−x) = f(x).

• Odd function: A function satisfying f(−x) = −f(x).

• Axis of symmetry: A line dividing a graph into mirror-image halves.

• Origin symmetry: Symmetry under a 180° rotation about the origin.

• Positive interval: An interval where f(x) > 0.

• Negative interval: An interval where f(x) < 0.

• Root or zero: An input value where f(x) = 0.

• Sign chart: A table showing where an expression is positive or negative.

• End behaviour: The behaviour of a function as its input approaches positive or negative infinity.

• Leading term: The term containing the highest power in a polynomial.

• Leading coefficient: The coefficient of the highest-power term.

• Degree: The highest exponent of the variable in a polynomial.

• Asymptote: A line that a graph approaches in a specified limiting direction.

• Horizontal asymptote: A horizontal line approached by a graph as x approaches positive or negative infinity.

• Vertical asymptote: A vertical line near which a function becomes unbounded on at least one side.

• Increasing interval: An interval where outputs rise as inputs increase.

• Decreasing interval: An interval where outputs fall as inputs increase.

• Extremum: A maximum or minimum value.

• Turning point: A point where a graph changes from increasing to decreasing or vice versa.

• Domain: The set of allowed input values.

• Range: The set of possible output values.

Key Takeaways

• Function features provide a systematic way to describe and compare graphs.

• Even functions have y-axis symmetry and satisfy f(−x) = f(x).

• Odd functions have origin symmetry and satisfy f(−x) = −f(x).

• A function may be neither even nor odd.

• Positive intervals occur where the graph lies above the x-axis.

• Negative intervals occur where the graph lies below the x-axis.

• Positive and negative intervals are different from increasing and decreasing intervals.

• End behaviour describes how a function behaves as x approaches positive or negative infinity.

• The leading term determines the end behaviour of a nonconstant polynomial.

• Exponential and rational functions may approach horizontal asymptotes.

• Periodic functions may continue oscillating rather than approaching a single value.

• Graphs can be compared using intercepts, symmetry, extrema, intervals, and end behaviour.

• Real-world graph analysis must consider the practical domain and meaning of the variables.

Check Your Understanding

Recall and Understanding

1. What is an even function? State the algebraic condition used to identify one.

2. What is an odd function? Describe its graphical symmetry.

3. Explain the difference between a positive interval and an increasing interval.

4. What is meant by the end behaviour of a function?

5. Why is the leading term important when describing the end behaviour of a polynomial?

Application

6. Determine whether f(x) = 4x⁴ − 3x² + 2 is even, odd, or neither.

7. Determine whether g(x) = 2x³ − 5x is even, odd, or neither.

8. Find the positive and negative intervals of f(x) = x² − 9.

9. Find the positive and negative intervals of g(x) = −x² + 16.

10. Describe the end behaviour of f(x) = 3x⁴ − 2x + 7.

11. Describe the end behaviour of g(x) = −2x⁵ + x³ − 4.

12. Determine the horizontal asymptote and end behaviour of h(x) = 3ˣ + 2.

Analysis and Challenge

13. Analyse f(x) = x² − 4x + 3. Identify its intercepts, axis of symmetry, minimum, positive and negative intervals, increasing and decreasing intervals, and end behaviour.

14. Compare the symmetry, extrema, and end behaviour of f(x) = x² and g(x) = −x³.

15. A student claims that f(x) = (x − 2)² is even because its graph has a vertical axis of symmetry. Explain the error.

Answers to Check Your Understanding

1. An even function has symmetry about the y-axis and satisfies:

f(−x) = f(x)

for every input in its domain.

2. An odd function has symmetry about the origin. Rotating its graph 180° around the origin produces the same graph.

Its algebraic condition is:

f(−x) = −f(x)

3. A positive interval is where f(x) > 0, meaning the graph is above the x-axis. An increasing interval is where the output rises as x increases.

4. End behaviour describes what happens to the outputs as x approaches positive or negative infinity.

5. For a polynomial, the leading term dominates the lower-degree terms when the magnitude of x becomes very large. Its degree and coefficient determine the graph's end behaviour.

6.

f(x) = 4x⁴ − 3x² + 2

f(−x) = 4(−x)⁴ − 3(−x)² + 2

f(−x) = 4x⁴ − 3x² + 2

Therefore:

The function is even.

7.

g(x) = 2x³ − 5x

g(−x) = −2x³ + 5x

g(−x) = −g(x)

Therefore:

The function is odd.

8.

f(x) = x² − 9

Factor:

f(x) = (x − 3)(x + 3)

Roots:

x = −3 and x = 3

Since the parabola opens upward:

Positive: (−∞, −3) and (3, ∞)

Negative: (−3, 3)

9.

g(x) = −x² + 16

Set g(x) = 0:

x² = 16

x = −4 or x = 4

The parabola opens downward.

Positive: (−4, 4)

Negative: (−∞, −4) and (4, ∞)

10.

f(x) = 3x⁴ − 2x + 7

The leading term is 3x⁴.

It has even degree and a positive leading coefficient.

Therefore:

As x → −∞, f(x) → ∞

As x → ∞, f(x) → ∞

Both ends rise.

11.

g(x) = −2x⁵ + x³ − 4

The leading term is −2x⁵.

It has odd degree and a negative leading coefficient.

Therefore:

As x → −∞, g(x) → ∞

As x → ∞, g(x) → −∞

The left end rises and the right end falls.

12.

h(x) = 3ˣ + 2

The parent exponential function is shifted upward by 2.

Horizontal asymptote:

y = 2

End behaviour:

As x → −∞, h(x) → 2.

As x → ∞, h(x) → ∞.

13.

f(x) = x² − 4x + 3

Factor:

f(x) = (x − 1)(x − 3)

x-intercepts:

(1, 0) and (3, 0)

y-intercept:

f(0) = 3

(0, 3)

Vertex x-coordinate:

x = −(−4)/(2 × 1) = 2

Vertex y-coordinate:

f(2) = 4 − 8 + 3 = −1

Minimum point:

(2, −1)

Axis of symmetry:

x = 2

Positive intervals:

(−∞, 1) and (3, ∞)

Negative interval:

(1, 3)

Decreasing interval:

(−∞, 2)

Increasing interval:

(2, ∞)

End behaviour:

Both ends rise toward positive infinity.

The function is neither even nor odd.

14.

For f(x) = x²:

• Even function.

• Symmetric about the y-axis.

• Minimum value 0 at x = 0.

• Both ends rise toward positive infinity.

For g(x) = −x³:

• Odd function.

• Symmetric about the origin.

• No local or absolute extrema over all real numbers.

• Left end rises toward positive infinity.

• Right end falls toward negative infinity.

Both functions pass through the origin, but their symmetry and end behaviour differ.

15.

The function:

f(x) = (x − 2)²

has a vertical axis of symmetry at:

x = 2

However, an even function must be symmetric specifically about the y-axis.

Check algebraically:

f(−x) = (−x − 2)² = (x + 2)²

This is not generally equal to (x − 2)².

Therefore:

The function is neither even nor odd.

Final Summary

Function features allow us to understand the overall behaviour of graphs by examining symmetry, positive and negative intervals, end behaviour, intercepts, extrema, and changes in direction.

By combining these characteristics, we can describe functions accurately, compare different mathematical relationships, and interpret graphs in real-world contexts.

A complete understanding of a graph comes from examining how its important features work together, rather than studying each feature separately.