Straight-line graphs
Coordinates are written (x, y) — along the corridor, then up the stairs — and work in all four quadrants, so both values can be negative.
Every straight line can be written as y = mx + c, where m is the gradient and c is the y-intercept.
Finding the equation of a line takes one of two routes.
Through one point with a given gradient: substitute m, then substitute the point's coordinates to find c.
Through two points: find the gradient first with m = (y₂ − y₁) ⁄ (x₂ − x₁), then use either point to find c.
Parallel lines have the same gradient. Perpendicular gradients multiply to −1, so each is the negative reciprocal of the other — flip the fraction and change the sign.
Worked example
Find the equation of the line perpendicular to y = 2x + 5 passing through (4, 3).
- The given line has gradient 2.
- Perpendicular gradients multiply to −1, so the new gradient is −1 ÷ 2 = −½.
- So the line is y = −½x + c.
- Substitute the point (4, 3): 3 = −½ × 4 + c
- That gives 3 = −2 + c, so c = 5.
Answer: y = −½x + 5. Check the gradients multiply to −1: 2 × (−½) = −1 ✓
- Gradient
- the steepness of a straight line, the m in y = mx + c, found as the change in y divided by the change in x
- y-intercept
- where a line crosses the y-axis, the c in y = mx + c
Quadratic graphs
A quadratic graph is a parabola, symmetrical about a vertical line through its turning point. It opens upwards when the x² coefficient is positive and downwards when it is negative.
Three features are examined:
• The roots — where the curve crosses the x-axis, which are the solutions of the equation when y = 0. Deduce them algebraically by factorising or by the quadratic formula. • The y-intercept — where it crosses the y-axis, found by substituting x = 0, which is simply the constant term. • The turning point — the minimum or maximum. At Higher tier this is found by completing the square: writing the expression as (x + p)² + q gives the turning point at (−p, q), read straight off.
The line of symmetry passes through the turning point, so it has equation x = −p. It also sits exactly midway between the two roots, which gives a quick way to find it when the roots are known.
- Quadratic graph
- a parabola, symmetrical about a vertical line through its turning point; it opens upwards when the x² coefficient is positive and downwards when it is negative
- Roots
- where a curve crosses the x-axis — the solutions of the equation when y = 0
- Turning point
- the maximum or minimum point of a curve, where it changes direction
Recognising and transforming graphs
Transformations of graphs are a Higher-tier requirement, and the rule that catches everyone is that changes inside the bracket behave in the opposite way to expectation.
• y = f(x) + a moves the graph up by a — outside the bracket, and it does what it looks like • y = f(x + a) moves the graph left by a — inside the bracket, and it moves the opposite way to the sign • y = −f(x) reflects the graph in the x-axis, turning it upside down • y = f(−x) reflects it in the y-axis, turning it left to right
So y = f(x − 3) moves the curve three to the right, not three to the left. The reason is worth a moment: to get the same output, x must now be 3 larger, so every point shifts right.
Common mistake
Inside the bracket, the movement is opposite to the sign, and it is horizontal. Outside the bracket, the movement matches the sign, and it is vertical. Deciding which it is comes first.
Real contexts, gradients and areas
Real-context graphs are read by asking what the gradient and the area actually mean, because that changes with the axes.
On a distance–time graph the gradient is the speed. A horizontal line means stationary, and a steeper line means faster.
On a velocity–time graph the gradient is the acceleration, and the area under the graph is the distance travelled. That pairing is the most examined idea in this part of the specification.
In financial contexts the gradient is a rate — cost per unit, or interest accruing per year.
On a curve the gradient changes constantly, so it is estimated at a point rather than calculated: draw a tangent — a straight line touching the curve at that point — and find the gradient of the tangent. It is an estimate, and answers will vary slightly, which the mark scheme allows for.
The area under a curve is estimated by splitting it into strips, usually trapezia, calculating each area and adding them. Using more, narrower strips gives a better estimate.
Whether an estimate is too big or too small is a standard follow-up question. If the trapezia sit above the curve, the estimate is too large; if the curve bulges above the straight tops, it is too small. Sketching one strip answers it.
None of this uses calculus, and none is expected to.
The equation of a circleHigher tier
A circle centred on the origin with radius r has equation
x² + y² = r²
which comes straight from Pythagoras: any point (x, y) on the circle is r from the origin. So x² + y² = 25 is a circle of radius 5 — take the square root of the number on the right, and do not mistake it for the radius itself.
Finding the tangent to a circle at a given point uses a circle theorem from Properties and constructions: the tangent is perpendicular to the radius at the point of contact. So the method is:
1. Find the gradient of the radius from the origin to the point, which is simply y ÷ x for that point 2. Take the negative reciprocal to get the tangent's gradient 3. Use that gradient and the point to find the equation, as in the straight-line section above
Worked example
Find the equation of the tangent to x² + y² = 25 at the point (3, 4).
- Check the point is on the circle: 3² + 4² = 9 + 16 = 25 ✓
- The radius runs from (0, 0) to (3, 4), so its gradient is 4 ÷ 3.
- The tangent is perpendicular to the radius, so its gradient is the negative reciprocal: −3⁄4.
- The tangent is y = −¾x + c, and it passes through (3, 4).
- Substitute: 4 = −¾ × 3 + c, so 4 = −2.25 + c, giving c = 6.25.
Answer: y = −¾x + 6.25, which can also be written 4y + 3x = 25. Every tangent-to-a-circle question follows these same three steps.
Before you practise
- y = mx + c: m is the gradient, c is the y-intercept. Rearrange into this form before reading either off.
- Gradient between two points = (y₂ − y₁) ⁄ (x₂ − x₁).
- Parallel lines share a gradient; perpendicular gradients multiply to −1.
- A quadratic's roots are where it crosses the x-axis; find them by factorising or the formula.
- Complete the square to get the turning point: (x + p)² + q has its turning point at (−p, q).
- Recognise linear, quadratic, cubic, reciprocal, exponential and trigonometric graphs by shape.
- f(x) + a moves up; f(x + a) moves left — inside the bracket is horizontal and opposite in sign.
- −f(x) reflects in the x-axis; f(−x) reflects in the y-axis.
- Distance–time gradient is speed; velocity–time gradient is acceleration and the area is distance.
- Estimate a gradient on a curve by drawing a tangent; estimate an area with strips, and say whether it is an over- or under-estimate.
- x² + y² = r² is a circle centred on the origin — the right-hand side is r², not r.
- A tangent is perpendicular to the radius at the point of contact.