AlgebraAQA GCSE Mathematics (8300)

Graphs

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Finding the equation of a straight line from two points
Foundation
Question
Find the equation of the line through $(-1, 2)$ and $(3, 10)$ .
Solution
1. 1
  Calculate the gradient using the two points.
  $m = \frac{10 - 2}{3 - (-1)} = \frac{8}{4} = 2$
2. 2
  Substitute one point and the gradient into $y = mx + c$ to find $c$ .
  $2 = 2(-1) + c \implies c = 4$
3. 3
  Write the final equation.
  $y = 2x + 4$
Answer
$y = 2x + 4$
Examiner note
Verify with the second point: $y = 2(3) + 4 = 10$ ✓
02.Finding the turning point by completing the square
Higher
Question
Find the coordinates of the turning point of $y = x^2 - 6x + 11$ .
Solution
1. 1
  Complete the square: halve the coefficient of $x$ and square it.
  $y = (x - 3)^2 - 9 + 11$
2. 2
  Simplify the constant terms.
  $y = (x - 3)^2 + 2$
3. 3
  Read off the turning point. The minimum occurs when the squared term is zero.
  $\text{Turning point: } (3, 2)$
Answer
Minimum turning point at $(3, 2)$
03.Estimating the gradient of a curve at a point
Higher
Question
Describe how to estimate the gradient of $y = x^2$ at the point $(2, 4)$ .
Solution
1. 1
  Draw a tangent to the curve at $(2, 4)$ — a straight line that just touches the curve at that point without crossing it.
2. 2
  Choose two points on the tangent that are far apart to minimise reading error. For example, the tangent to $y = x^2$ at $(2,4)$ passes through $(0, 0)$ and $(3, 6)$ .
  $m \approx \frac{6 - 0}{3 - 0} = 2$
3. 3
  The gradient of the curve at $x = 2$ is approximately 4 (the exact calculus result). The tangent gives a close estimate depending on accuracy of the sketch.
Answer
Gradient $\approx 4$ at $x = 2$ (estimate from tangent)
Examiner note
Marks are awarded for the method (drawing a tangent, reading two points, calculating rise/run) even if the numerical answer differs slightly.
04.Finding the equation of a tangent to a circle
Higher
Question
Find the equation of the tangent to $x^2 + y^2 = 50$ at the point $(5, 5)$ .
Solution
1. 1
  Verify the point lies on the circle.
  $5^2 + 5^2 = 25 + 25 = 50 \checkmark$
2. 2
  Find the gradient of the radius from origin to $(5, 5)$ .
  $m_{\text{radius}} = \frac{5}{5} = 1$
3. 3
  The tangent is perpendicular to the radius.
  $m_{\text{tangent}} = -1$
4. 4
  Find the equation of the tangent using point-slope form.
  $y - 5 = -1(x - 5) \implies y = -x + 10$
Answer
$y = -x + 10$ (or equivalently $x + y = 10$ )

Key formulae

Equation of a straight line
$y = mx + c$
- $m$
  — gradient (rise ÷ run)
- $c$
  — y-intercept
When to use
To write, read or rearrange the equation of any non-vertical straight line.
Example
Gradient 3, y-intercept −2: $y = 3x - 2$
Gradient between two points
$m = \frac{y_2 - y_1}{x_2 - x_1}$
- $(x_1, y_1)$
  — first point
- $(x_2, y_2)$
  — second point
When to use
To find the gradient when two points on the line are known.
Example
Through $(1,3)$ and $(4,9)$ : $m = (9-3)/(4-1) = 2$
Quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
- $a$
  — coefficient of $x^2$
- $b$
  — coefficient of $x$
- $c$
  — constant term
When to use
To find the roots of $ax^2 + bx + c = 0$ when factorising is not obvious.
Example
Provided on the AQA formula sheet — apply it to any quadratic equation
Equation of a circle (centred at origin)
$x^2 + y^2 = r^2$
- $r$
  — radius of the circle
When to use
To identify or write the equation of a circle centred at the origin.
Example
Radius 6: $x^2 + y^2 = 36$

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Gradient: A measure of the steepness of a line or curve. For a straight line, gradient = rise ÷ run. For a curve, it is the gradient of the tangent at that point.
Root (of a graph): The x-value(s) where the graph crosses the x-axis, i.e. where y = 0. A quadratic can have 0, 1 or 2 roots.
Turning point: The point on a curve where the gradient changes from positive to negative (maximum) or negative to positive (minimum). Found at x = −b/(2a) for a quadratic.
Asymptote: A line that a curve approaches but never reaches. For y = 1/x, there are asymptotes at x = 0 and y = 0; for y = kˣ, there is an asymptote at y = 0.
Tangent (to a curve): A straight line that touches a curve at exactly one point without crossing it. Its gradient equals the gradient of the curve at that point.
Discriminant: The expression b² − 4ac from the quadratic formula. It determines the number of real roots: positive → two roots; zero → one repeated root; negative → no real roots.

Common mistakes and misconceptions

Mistake
Reading gradient as run/rise instead of rise/run
Why it happens
Students confuse the order of subtraction or mix up which axis represents which direction.
How to avoid it
Gradient = (change in y) ÷ (change in x). Always divide the vertical change by the horizontal change, and check the sign: upward slope → positive; downward slope → negative.
Mistake
Thinking $f(x + 3)$ shifts the graph right
Why it happens
The $+3$ inside the bracket feels like 'adding 3 to x', so students expect a rightward shift.
How to avoid it
$f(x + 3)$ shifts left by 3 because the same y-value now occurs at an x-value that is 3 less. Remember: inside the bracket, the shift is opposite to the sign.
Mistake
Squaring the radius instead of the whole equation: writing $x^2 + y^2 = r$ for a circle
Why it happens
Students forget to square $r$ in the circle equation.
How to avoid it
The equation is always $x^2 + y^2 = r^2$ . For radius 5, this is $x^2 + y^2 = 25$ , not 5.
Mistake
Drawing a chord instead of a tangent when estimating gradient
Why it happens
A chord connects two points on the curve; a tangent just touches it at one point — students mix them up.
How to avoid it
A tangent must touch the curve at only one point and not cut through it. Use a ruler to draw the steepest or shallowest line that just grazes the curve at the required point.

AlgebraAQA GCSE Mathematics (8300)

Graphs

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

Previous Next

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Finding the equation of a straight line from two points
Foundation
Question
Find the equation of the line through $(-1, 2)$ and $(3, 10)$ .
Solution
1. 1
  Calculate the gradient using the two points.
  $m = \frac{10 - 2}{3 - (-1)} = \frac{8}{4} = 2$
2. 2
  Substitute one point and the gradient into $y = mx + c$ to find $c$ .
  $2 = 2(-1) + c \implies c = 4$
3. 3
  Write the final equation.
  $y = 2x + 4$
Answer
$y = 2x + 4$
Examiner note
Verify with the second point: $y = 2(3) + 4 = 10$ ✓
02.Finding the turning point by completing the square
Higher
Question
Find the coordinates of the turning point of $y = x^2 - 6x + 11$ .
Solution
1. 1
  Complete the square: halve the coefficient of $x$ and square it.
  $y = (x - 3)^2 - 9 + 11$
2. 2
  Simplify the constant terms.
  $y = (x - 3)^2 + 2$
3. 3
  Read off the turning point. The minimum occurs when the squared term is zero.
  $\text{Turning point: } (3, 2)$
Answer
Minimum turning point at $(3, 2)$
03.Estimating the gradient of a curve at a point
Higher
Question
Describe how to estimate the gradient of $y = x^2$ at the point $(2, 4)$ .
Solution
1. 1
  Draw a tangent to the curve at $(2, 4)$ — a straight line that just touches the curve at that point without crossing it.
2. 2
  Choose two points on the tangent that are far apart to minimise reading error. For example, the tangent to $y = x^2$ at $(2,4)$ passes through $(0, 0)$ and $(3, 6)$ .
  $m \approx \frac{6 - 0}{3 - 0} = 2$
3. 3
  The gradient of the curve at $x = 2$ is approximately 4 (the exact calculus result). The tangent gives a close estimate depending on accuracy of the sketch.
Answer
Gradient $\approx 4$ at $x = 2$ (estimate from tangent)
Examiner note
Marks are awarded for the method (drawing a tangent, reading two points, calculating rise/run) even if the numerical answer differs slightly.
04.Finding the equation of a tangent to a circle
Higher
Question
Find the equation of the tangent to $x^2 + y^2 = 50$ at the point $(5, 5)$ .
Solution
1. 1
  Verify the point lies on the circle.
  $5^2 + 5^2 = 25 + 25 = 50 \checkmark$
2. 2
  Find the gradient of the radius from origin to $(5, 5)$ .
  $m_{\text{radius}} = \frac{5}{5} = 1$
3. 3
  The tangent is perpendicular to the radius.
  $m_{\text{tangent}} = -1$
4. 4
  Find the equation of the tangent using point-slope form.
  $y - 5 = -1(x - 5) \implies y = -x + 10$
Answer
$y = -x + 10$ (or equivalently $x + y = 10$ )

Key formulae

Equation of a straight line
$y = mx + c$
- $m$
  — gradient (rise ÷ run)
- $c$
  — y-intercept
When to use
To write, read or rearrange the equation of any non-vertical straight line.
Example
Gradient 3, y-intercept −2: $y = 3x - 2$
Gradient between two points
$m = \frac{y_2 - y_1}{x_2 - x_1}$
- $(x_1, y_1)$
  — first point
- $(x_2, y_2)$
  — second point
When to use
To find the gradient when two points on the line are known.
Example
Through $(1,3)$ and $(4,9)$ : $m = (9-3)/(4-1) = 2$
Quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
- $a$
  — coefficient of $x^2$
- $b$
  — coefficient of $x$
- $c$
  — constant term
When to use
To find the roots of $ax^2 + bx + c = 0$ when factorising is not obvious.
Example
Provided on the AQA formula sheet — apply it to any quadratic equation
Equation of a circle (centred at origin)
$x^2 + y^2 = r^2$
- $r$
  — radius of the circle
When to use
To identify or write the equation of a circle centred at the origin.
Example
Radius 6: $x^2 + y^2 = 36$

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Gradient: A measure of the steepness of a line or curve. For a straight line, gradient = rise ÷ run. For a curve, it is the gradient of the tangent at that point.
Root (of a graph): The x-value(s) where the graph crosses the x-axis, i.e. where y = 0. A quadratic can have 0, 1 or 2 roots.
Turning point: The point on a curve where the gradient changes from positive to negative (maximum) or negative to positive (minimum). Found at x = −b/(2a) for a quadratic.
Asymptote: A line that a curve approaches but never reaches. For y = 1/x, there are asymptotes at x = 0 and y = 0; for y = kˣ, there is an asymptote at y = 0.
Tangent (to a curve): A straight line that touches a curve at exactly one point without crossing it. Its gradient equals the gradient of the curve at that point.
Discriminant: The expression b² − 4ac from the quadratic formula. It determines the number of real roots: positive → two roots; zero → one repeated root; negative → no real roots.

Common mistakes and misconceptions

Mistake
Reading gradient as run/rise instead of rise/run
Why it happens
Students confuse the order of subtraction or mix up which axis represents which direction.
How to avoid it
Gradient = (change in y) ÷ (change in x). Always divide the vertical change by the horizontal change, and check the sign: upward slope → positive; downward slope → negative.
Mistake
Thinking $f(x + 3)$ shifts the graph right
Why it happens
The $+3$ inside the bracket feels like 'adding 3 to x', so students expect a rightward shift.
How to avoid it
$f(x + 3)$ shifts left by 3 because the same y-value now occurs at an x-value that is 3 less. Remember: inside the bracket, the shift is opposite to the sign.
Mistake
Squaring the radius instead of the whole equation: writing $x^2 + y^2 = r$ for a circle
Why it happens
Students forget to square $r$ in the circle equation.
How to avoid it
The equation is always $x^2 + y^2 = r^2$ . For radius 5, this is $x^2 + y^2 = 25$ , not 5.
Mistake
Drawing a chord instead of a tangent when estimating gradient
Why it happens
A chord connects two points on the curve; a tangent just touches it at one point — students mix them up.
How to avoid it
A tangent must touch the curve at only one point and not cut through it. Use a ruler to draw the steepest or shallowest line that just grazes the curve at the required point.

Graphs

Master the topic

Step-by-step worked examples

01.Finding the equation of a straight line from two points

02.Finding the turning point by completing the square

03.Estimating the gradient of a curve at a point

04.Finding the equation of a tangent to a circle

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Graphs

Master the topic

Step-by-step worked examples

01.Finding the equation of a straight line from two points

02.Finding the turning point by completing the square

03.Estimating the gradient of a curve at a point

04.Finding the equation of a tangent to a circle

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions