Lesson 7 ~30 min Unit 2 · Linear Relationships +85 XP

Positive and Negative Gradients

Master the four types of gradient — positive, negative, zero and undefined. Learn to read gradient sign from graphs, tables and equations.

Today's hook: A hiker walks up a hill, then down the other side. The gradient going up is positive — going down is negative. What about when they reach the peak?

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Think First

warm-up

A hiker walks up a hill, then down the other side. Sketch a graph of height vs distance. What is the gradient going up compared to going down?

Reveal answer

Going up: As distance increases, height increases. Gradient is positive. Going down: As distance increases, height decreases. Gradient is negative. At the top: The hill is momentarily flat. Gradient is zero.

Positive Gradients

+5 XP

A line has a positive gradient when it goes uphill from left to right ($m > 0$). As $x$ increases, $y$ also increases.

For a positive gradient: the line rises left-to-right. Examples: $y = 2x + 1$ ($m = 2$), $y = \frac{1}{2}x - 3$ ($m = \frac{1}{2}$), $y = 5x$ ($m = 5$). All slope uphill.

$m > 0$ → rises left-to-right → $y$ increases as $x$ increases

Watch Me Solve It · 3 examples

Watch Me Solve It · Sketching $y = 3x + 1$

+15 XP per step

PROBLEM

Sketch $y = 3x + 1$ and explain why it has a positive gradient.

1

Read the equation

Gradient $m = 3$ (positive), $y$-intercept $c = 1$

The equation is in the form $y = mx + c$. The coefficient of $x$ is the gradient.
2

Plot two points

Plot $(0, 1)$. From rise = 3, run = 1, find $(1, 4)$.
3

Draw and explain

Draw the line through $(0,1)$ and $(1,4)$ — it slopes uphill left-to-right

$m = 3 > 0$, so the line is positive. It rises 3 units for every 1 unit across.

AnswerPositive gradient: $m = 3 > 0$. Line rises steeply left-to-right.

Negative Gradients

+5 XP

A line has a negative gradient when it goes downhill from left to right ($m < 0$). As $x$ increases, $y$ decreases.

For a negative gradient: the line falls left-to-right. The rise is negative (going down). Examples: $y = -2x + 5$ ($m = -2$), $y = -\frac{1}{3}x + 4$ ($m = -\frac{1}{3}$), $y = -x$ ($m = -1$).

$m < 0$ → falls left-to-right → $y$ decreases as $x$ increases

Watch Me Solve It · Sketching $y = -2x + 4$

+15 XP per step

PROBLEM

Sketch $y = -2x + 4$ and explain why it has a negative gradient.

1

Read the equation

Gradient $m = -2$ (negative), $y$-intercept $c = 4$
2

Plot two points

Plot $(0,4)$. Rise $= -2$, run $= 1$: go right 1, down 2 to $(1,2)$.
3

Draw and explain

Draw through $(0,4)$ and $(1,2)$ — it slopes downhill left-to-right

$m = -2 < 0$, so the gradient is negative. For every 1 unit across, the line falls 2 units.

AnswerNegative gradient: $m = -2 < 0$. Line falls steeply left-to-right.

Zero and Undefined Gradient

+5 XP

Zero ($m = 0$) — Horizontal

Line is perfectly flat, parallel to $x$-axis. $y$ stays constant as $x$ changes. Equation: $y = c$. Example: $y = 5$ has gradient 0.

Undefined — Vertical

Line is straight up and down, parallel to $y$-axis. Run $= 0$, so $m = \dfrac{\text{rise}}{0}$ is impossible. Equation: $x = a$. Example: $x = 4$ has undefined gradient.

Common confusion: Zero and undefined are not the same. Zero = flat horizontal. Undefined = straight vertical. They are opposites.

Memory trick: Vertical = Undefined. The run is zero — division by zero has no value.

Watch Me Solve It · Predicting gradient sign from a table

+15 XP per step

PROBLEM

Determine the gradient sign for this table: $x$ = 0, 1, 2, 3, 4, 5 and $y$ = 12, 9, 6, 3, 0, −3.

1

Observe the direction of $y$

$x$ increases: $0 \to 1 \to 2 \to 3 \to 4 \to 5$

We read left-to-right, so $x$ always increases.
2

Check what $y$ does

$y$ decreases: $12 \to 9 \to 6 \to 3 \to 0 \to -3$

$y$ goes down, so the gradient must be negative.

AnswerNegative gradient ($m = -3$). $y$ decreases as $x$ increases.

Predicting Gradient Sign from Tables

key rule

To predict gradient sign from a table, look only at the $y$-column direction as $x$ increases:

Positive ($m > 0$)

As $x$ increases, $y$ also increases. E.g., $y$: 3, 5, 7, 9 → $m > 0$

Negative ($m < 0$)

As $x$ increases, $y$ decreases. E.g., $y$: 10, 7, 4, 1 → $m < 0$

Zero ($m = 0$)

$y$ stays the same as $x$ increases. E.g., $y$: 6, 6, 6, 6 → $m = 0$

Watch out

All positive $y$-values does NOT mean positive gradient. $y = 10, 8, 6, 4$ has negative gradient even though all $y$-values are positive.

Summary: All Four Gradient Types

reference

Positive$m > 0$. Uphill. $y = mx + c$ with $m > 0$.

Negative$m < 0$. Downhill. $y = mx + c$ with $m < 0$.

Zero$m = 0$. Horizontal. Equation: $y = c$.

UndefinedVertical. Run = 0. Equation: $x = a$.

Reading Gradient from an Equation

+5 XP

For equations in the form $y = mx + c$, the gradient is the coefficient of $x$. Read its sign directly:

$y = 4x - 3$ → $m = 4$ (positive)

The coefficient of $x$ is 4. Positive coefficient = uphill.

$y = -3x + 2$ → $m = -3$ (negative)

The coefficient of $x$ is $-3$. Negative coefficient = downhill.

$y = 7$ → $m = 0$ (zero)

No $x$ term means $m = 0$. Horizontal line.

$x = 4$ → undefined gradient

$x = \text{constant}$ means vertical line. Undefined.

Common Pitfalls

heads-up

Saying "going down" without specifying left to right

A line that goes down from right to left would be positive when read in the standard left-to-right direction.

Fix: Always state "goes downhill from left to right". The convention is always left-to-right.

Thinking $m = 0$ means "no line"

Zero gradient just means the line is horizontal. The line exists perfectly well — it just doesn't slope.

Fix: Zero gradient = horizontal line $y = c$. Still a real line.

Predicting from positive $y$-values alone

$y$ going from 10 to 8 to 6 has negative gradient, even though all values are positive.

Fix: Look at the direction of change in $y$, not the sign of $y$ values.

Negative gradient = line is below $x$-axis

A negative gradient means downhill direction, not position. $y = -x + 10$ has negative gradient but is well above the $x$-axis at $x = 0$.

Fix: Gradient tells us direction. The $y$-intercept determines position.

Copy Into Your Books

Positive ($m > 0$)

Uphill left-to-right
As $x$ increases, $y$ increases
Example: $y = 2x + 1$

Negative ($m < 0$)

Downhill left-to-right
As $x$ increases, $y$ decreases
Example: $y = -3x + 2$

Zero ($m = 0$)

Flat horizontal line
$y$ stays constant
Equation: $y = c$

Undefined

Straight vertical line
Run = 0, division by zero
Equation: $x = a$

Brain Trainer · 10 problems

Brain Trainer · Gradient Sign Drills

10 problems

Quick-fire questions. Answer each, then reveal.

1 Does $y = 4x - 3$ have a positive or negative gradient?

Positive. $m = 4 > 0$.Positive
2 What is the gradient of $y = 7$?

$m = 0$. Horizontal line.0
3 What is the gradient of $x = -2$?

Undefined. Vertical line; run = 0.Undefined
4 $y$-values: $8, 6, 4, 2, 0$. Positive, negative, or zero gradient?

Negative. $y$ decreases; $m = -2$.Negative
5 Describe a line with $m = -1$ through $(0, 3)$.

Through $(0,3)$, slopes downhill at 45°. Points: $(1,2)$, $(-1,4)$.Downhill at 45°, through (0,3)
6 Does $y = -\frac{1}{2}x + 10$ have positive or negative gradient?

Negative. $m = -\frac{1}{2} < 0$. Gently downhill.Negative
7 All points on a line have $y = -3$. What is the gradient?

$m = 0$. Horizontal line $y = -3$.0
8 A parked car's distance-time graph: what is the gradient?

$m = 0$. Distance doesn't change; horizontal line.0
9 Line through $(1,2)$ and $(4,8)$: positive or negative gradient?

Positive. Both $x$ and $y$ increase. $m = \frac{6}{3} = 2$.Positive
10 Name the four gradient types and give an example equation for each.

(1) Positive: $y = 3x + 1$. (2) Negative: $y = -2x + 5$. (3) Zero: $y = 4$. (4) Undefined: $x = 3$.Positive, Negative, Zero, Undefined

Quick Check · 5 questions

A line going uphill from left to right has:

+10 XP

A table shows $x$ increasing and $y$ decreasing. The gradient type is:

+10 XP

Which equation has a zero gradient?

+10 XP

A line with negative gradient through $(0, 2)$ goes:

+10 XP

What is the gradient of the line $x = 4$?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Medium 3 MARKS

SAQ 1. Sketch four lines on one set of axes, all passing through $(0,0)$, with gradients $m = 2$, $m = -1$, $m = 0$, and undefined (vertical). Label each with its gradient and equation.

Sketch in your workbook.

Apply Easy 3 MARKS

SAQ 2. A car's distance from home decreases over time as it drives back. Describe the gradient of the distance-time graph and explain your reasoning.

Answer in your workbook.

Reason Medium 3 MARKS

SAQ 3. Without drawing, determine whether $y = -3x + 2$ has a positive or negative gradient. Explain how you know.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — Positive gradient: uphill left-to-right, $m > 0$.

2. B — Negative: $y$ decreases as $x$ increases.

3. C — $y = 4$ is horizontal with zero gradient.

4. B — Negative gradient means downhill left-to-right.

5. D — $x = 4$ is vertical; undefined gradient (run = 0).

Model Answers

SAQ 1: $m = 2$ → steep uphill, equation $y = 2x$. $m = -1$ → downhill 45°, equation $y = -x$. $m = 0$ → horizontal, equation $y = 0$. Undefined → vertical, equation $x = 0$.

SAQ 2: Negative gradient. As time increases ($x$ increases), distance decreases ($y$ decreases). Since $y$ decreases as $x$ increases, $m < 0$. Line slopes downhill left-to-right. Example: $d = -30t + 60$ has gradient $-30$.

SAQ 3: Negative gradient. In $y = -3x + 2$, the coefficient of $x$ is $-3$. Since $-3 < 0$, the gradient is negative. The line goes downhill from left to right. Check: at $x = 0$, $y = 2$; at $x = 1$, $y = -1$. The $y$-value decreased, confirming $m < 0$.

Stretch Challenge · +25 XP, +10 coins

Gradient Sign Challenges

Part A: A line passes through $(2, 5)$ and has gradient $-2$. Find two more points on this line.

Part B: A student says "if a line has negative gradient, it means the line is below the $x$-axis." Find a counterexample to show this is false.

Part C: The points $(1, 4)$, $(3, k)$ and $(5, 0)$ are collinear. Find $k$ and state the gradient type.

Reveal solutions

Part A: From $(2,5)$ with $m = -2$: forward $(2+1, 5-2) = (3,3)$. Backward $(2-1, 5+2) = (1,7)$.

Part B: $y = -x + 10$ has $m = -1$ (negative), but at $x = 0$, $y = 10$ — well above the $x$-axis. Gradient tells direction, not position.

Part C: $m = \frac{0-4}{5-1} = -1$ (negative). Using $m = -1$ from $(1,4)$ to $(3,k)$: $-1 = \frac{k-4}{2}$, so $k = 2$.

Quick Review

Positive ($m > 0$)

Uphill left-to-right. $y$ increases as $x$ increases.

Negative ($m < 0$)

Downhill left-to-right. $y$ decreases as $x$ increases.

Zero ($m = 0$)

Flat horizontal. $y$ stays constant. Equation $y = c$.

Undefined

Vertical. Run = 0. Equation $x = a$. Division by zero.

From a table

$y$ up = positive; $y$ down = negative; $y$ same = zero.

From an equation

Coefficient of $x$ = gradient. Read its sign.

Mark lesson as complete

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Unit overview · Maths subject page