Open Math›Year 10›Non-linear relationships›Comparing relationships

Comparing relationships

Identify relationship types from tables of values, match equations to graph types, find intersection points of linear and non-linear graphs, and compare rates of growth.

Worked examples

Identifying a relationship type from a table

Straightforward

Problem

For each table below, identify whether the relationship is linear, quadratic or exponential and state the key feature used.

(a)

\begin{array}{c|cccc} x & 1 & 2 & 3 & 4 \\ \hline y & 4 & 7 & 10 & 13 \end{array}

(b)

\begin{array}{c|cccc} x & 0 & 1 & 2 & 3 \\ \hline y & 5 & 10 & 20 & 40 \end{array}

For table (a), find the first differences between consecutive

y

-values.

First differences:

7-4=3

10-7=3

13-10=3

. All equal to 3.

Classify the relationship in table (a).

The first differences are constant (all equal to 3), so the relationship is linear. The gradient is 3 and the equation is

y = 3x + 1

For table (b), check the first differences, then the ratio of consecutive

y

-values.

First differences:

10-5=5

20-10=10

40-20=20

. Not constant, so not linear.

Ratios:

\dfrac{10}{5}=2

\dfrac{20}{10}=2

\dfrac{40}{20}=2

. The ratio is constant (each

y

-value doubles).

Classify the relationship in table (b).

A constant multiplicative ratio is the defining feature of an exponential relationship. The equation is

y = 5 \times 2^x

Answer

(a) Linear — first differences are constant (all equal to 3). (b) Exponential — the ratio of consecutive

y

-values is constant (all equal to 2).

Matching equations to relationship types and explaining identifying features

Moderate

Problem

For each equation, state the type of relationship and give one identifying feature.

(a)

y = 3x - 4

(b)

y = 2x^2 + 1

(c)

y = 5 \times (0.8)^x

Classify

y = 3x - 4

This is in the form

y = mx + c

with

m = 3

and

c = -4

. The variable

x

appears only to the power of 1. This is a linear relationship — its graph is a straight line and first differences in a table would all equal 3.

Classify

y = 2x^2 + 1

The highest power of

x

is 2. This is a quadratic relationship — its graph is a parabola opening upward. In a table, the second differences would all be constant (equal to

2 \times 2 = 4

, since the coefficient of

x^2

is 2).

Classify

y = 5 \times (0.8)^x

The variable

x

appears as the exponent. The base is

0.8

, and since

0 < 0.8 < 1

, this is exponential decay. The

y

-intercept is

5 \times (0.8)^0 = 5

. In a table, the ratio of consecutive

y

-values would all equal 0.8. The graph approaches

y = 0

x

increases.

Answer

(a) Linear — straight-line graph with gradient 3. (b) Quadratic — parabola; second differences are constant. (c) Exponential decay — base

0.8

satisfies

0 < 0.8 < 1

; ratio of consecutive

y

-values equals 0.8.

Finding the intersection of a linear and quadratic graph and comparing growth

Challenging

Problem

The table below shows values of

y = 2x

and

y = x^2

for

x = 0

5

\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y=2x & 0 & 2 & 4 & 6 & 8 & 10 \\ y=x^2 & 0 & 1 & 4 & 9 & 16 & 25 \end{array}

(a) Find the coordinates of the positive intersection point.
(b) State which function gives larger

y

-values for

x > 2

Read the table to find where both functions give the same

y

-value for positive

x

x = 0

y = 2(0)=0

and

y = 0^2=0

. Equal, but

x=0

is not positive.
At

x = 2

y = 2(2)=4

and

y = 2^2=4

. Equal — this is the positive intersection.

Write the coordinates of the positive intersection point.

The positive intersection point is

(2, 4)

Verify by solving algebraically.

Setting

2x = x^2

x^2 - 2x = 0 \Rightarrow x(x-2)=0

. Solutions:

x = 0

x = 2

. This confirms the positive intersection at

x = 2

Compare the functions for

x > 2

x = 3

2(3)=6

and

3^2=9

. Quadratic is larger.
At

x = 4

8

and

16

. Quadratic is larger.
At

x = 5

10

and

25

. Quadratic is larger and the gap is growing.

For

x > 2

y = x^2

gives larger

y

-values. Quadratic growth is faster than linear growth for large

x

Answer

(a) The positive intersection is at

(2, 4)

. (b) For

x > 2

y = x^2

gives larger

y

-values because quadratic growth eventually outpaces linear growth.

Practise

Q1·Straightforward

The table below shows a relationship between

x

and

y

\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 3 & 5 & 7 & 9 & 11 \end{array}

Which type of relationship does the table show?

LinearQuadraticExponentialNone of these

Explanation

First differences:

5-3=2

7-5=2

9-7=2

11-9=2

. All equal 2, so the first differences are constant. Constant first differences is the defining feature of a linear relationship.

Q2·Straightforward

The table below shows a relationship between

x

and

y

\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 0 & 1 & 4 & 9 & 16 \end{array}

Which type of relationship does the table show?

LinearQuadraticExponentialNone of these

Explanation

First differences:

1-0=1

4-1=3

9-4=5

16-9=7

. Not constant, so not linear. Second differences:

3-1=2

5-3=2

7-5=2

. The second differences are constant (all equal to 2), confirming a quadratic relationship. (The equation is

y = x^2

Q3·Straightforward

The table below shows a relationship between

x

and

y

\begin{array}{c|cccc} x & 0 & 1 & 2 & 3 \\ \hline y & 2 & 6 & 18 & 54 \end{array}

Which type of relationship does the table show?

LinearQuadraticExponentialNone of these

Explanation

Ratios:

\dfrac{6}{2}=3

\dfrac{18}{6}=3

\dfrac{54}{18}=3

. Each

y

-value is multiplied by 3 when

x

increases by 1. A constant multiplicative ratio identifies an exponential relationship.

Q4·Straightforward

The table below shows a relationship between

x

and

y

\begin{array}{c|ccccc} x & 1 & 2 & 3 & 4 & 5 \\ \hline y & 2 & 5 & 10 & 17 & 26 \end{array}

Which type of relationship does the table show?

LinearQuadraticExponentialNone of these

Explanation

First differences:

5-2=3

10-5=5

17-10=7

26-17=9

. Not constant, so not linear. Second differences:

5-3=2

7-5=2

9-7=2

. The second differences are constant (all equal to 2), confirming a quadratic relationship. (The equation is

y = x^2 + 1

Q5·Moderate

Which of the following equations represents an exponential relationship?

$y = 2x + 1$$y = x^2 + 3$$y = 4 \times 3^x$$y = -5x + 2$

Explanation

y = 4 \times 3^x

, the variable

x

appears as the exponent. This is an exponential relationship. The other options are linear (

y = 2x + 1

and

y = -5x + 2

) and quadratic (

y = x^2 + 3

), where

x

is not an exponent.

Q6·Moderate

Which feature confirms that a table of values shows a linear relationship?

The ratio of consecutive $y$-values is constant.The first differences between consecutive $y$-values are constant.The second differences between consecutive $y$-values are constant.The $y$-values are always positive.

Explanation

A constant rate of change (gradient) means

y

increases or decreases by the same fixed amount each time

x

increases by 1. In a table, this appears as constant first differences. A constant ratio identifies an exponential relationship; constant second differences identify a quadratic relationship.

Q7·Moderate

The table below shows a relationship between

x

and

y

\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 1 & 2 & 4 & 8 & 16 \end{array}

Which type of relationship does the table show, and what feature confirms this?

Linear, because the $y$-values increase by the same amount each timeQuadratic, because the second differences are equalExponential, because each $y$-value is double the previous oneNone of these

Explanation

Ratios:

\dfrac{2}{1}=2

\dfrac{4}{2}=2

\dfrac{8}{4}=2

\dfrac{16}{8}=2

. The ratio is constant (equal to 2), confirming an exponential relationship. First differences (1, 2, 4, 8) are not constant, ruling out linear. Second differences (1, 2, 4) are not constant, ruling out quadratic. The equation is

y = 2^x

Q8·Moderate

The graphs of

y = 3x + 2

and

y = x^2

are drawn on the same axes. For large positive values of

x

, which function produces the greater

y

-values?

$y = 3x + 2$, because it has a constant gradient$y = x^2$, because quadratic functions grow faster than linear functions for large $x$Both produce equal $y$-values for all large $x$$y = 3x + 2$, because it has a larger $y$-intercept

Explanation

x = 10

y = 3(10)+2=32

and

y = 10^2=100

. At

x = 100

y = 302

and

y = 10000

. For large

x

y = x^2

grows much faster. Although

y = 3x + 2

is larger for small positive

x

(at

x = 1

5 > 1

), the quadratic eventually overtakes it.

Q9·Challenging

The table below shows values of

y = 4x

and

y = 2^x

\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y=4x & 0 & 4 & 8 & 12 & 16 & 20 \\ y=2^x & 1 & 2 & 4 & 8 & 16 & 32 \end{array}

Find the coordinates of the point where the two graphs intersect.

Explanation

Reading the table: when

x = 4

, both

y = 4 \times 4 = 16

and

y = 2^4 = 16

. The two graphs meet at the point

(4, 16)

Q10·Challenging

The table below shows values of

y = 4x

and

y = 2^x

\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y=4x & 0 & 4 & 8 & 12 & 16 & 20 \\ y=2^x & 1 & 2 & 4 & 8 & 16 & 32 \end{array}

For

x > 4

, which function gives the larger

y

-value? Select the correct answer and reason.

$y = 4x$, because linear functions have a constant gradient$y = 2^x$, because exponential functions eventually grow faster than linear functionsBoth functions give equal $y$-values for all $x > 4$$y = 4x$, because the coefficient 4 is greater than the base 2

Explanation

x = 5

y = 4(5) = 20

and

y = 2^5 = 32

, so

y = 2^x

is already larger. As

x

continues to increase,

2^x

doubles at each step while

4x

only increases by 4. Exponential growth will always eventually outpace linear growth, regardless of the size of the linear coefficient.

Q11·Challenging

The table below shows values of

y = 3x

and

y = x^2

\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y=3x & 0 & 3 & 6 & 9 & 12 & 15 \\ y=x^2 & 0 & 1 & 4 & 9 & 16 & 25 \end{array}

Find the coordinates of the positive intersection point of the two graphs.

Explanation

x = 0

, both functions equal 0, but

x = 0

is not positive. At

x = 3

y = 3 \times 3 = 9

and

y = 3^2 = 9

. Both give

y = 9

, so the positive intersection point is

(3, 9)

Q12·Challenging

The table below shows values of

y = 3x

and

y = x^2

\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y=3x & 0 & 3 & 6 & 9 & 12 & 15 \\ y=x^2 & 0 & 1 & 4 & 9 & 16 & 25 \end{array}

For

x > 3

, which function gives the larger

y

-value, and what does this tell us about the rates of growth?

$y = 3x$, because it has a gradient of 3$y = x^2$, because quadratic growth eventually outpaces linear growthBoth give equal $y$-values for all $x > 3$$y = 3x$, because the coefficient 3 is always larger than $x$ for $x < 3$

Explanation

x = 4

y = 3(4)=12

and

y = 4^2=16

, so

y = x^2

is larger. At

x = 5

y = 15

and

y = 25

. The gap keeps widening. Although

y = 3x > y = x^2

for

0 < x < 3

, the quadratic overtakes the linear at

x = 3

and stays ahead. Quadratic growth is faster than linear growth for all sufficiently large values of

x