AQA GCSE Maths · 3.2.2

Graphs

50 maths questions with worked answers with answers for 3.2.2 Graphs in AQA GCSE Maths (spec 8300). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.

All 50 questions and answers

Question 1 · Coordinates · A8
In which quadrant is the point (−3, 4)?
Answer:
The x-coordinate is negative and the y-coordinate is positive.
That places (−3, 4) in quadrant II.
Question 2 · Coordinates · A8
Give the coordinates of the origin.
Answer:
The origin is where the x-axis and y-axis meet.
Both coordinates are zero: (0, 0).
Question 3 · Coordinates · A8
What is the x-coordinate of (7, −2)?
Answer:
Coordinates are written in the order (x, y).
The first coordinate is 7.
Question 4 · Coordinates · A8
A point moves 3 right and 2 down from (1, 5). Where does it end?
Answer:
Add 3 to the x-coordinate: 1 + 3 = 4.
Subtract 2 from the y-coordinate: 5 − 2 = 3.
The new point is (4, 3).
Question 5 · Straight lines · A9
Find y when y = 2x + 3 and x = 4.
Answer:
Substitute x = 4: y = 2 × 4 + 3.
y = 8 + 3 = 11.
Question 6 · Straight lines · A10
State the gradient of y = 5x − 2.
Answer:
Compare y = 5x − 2 with y = mx + c.
The gradient m is 5.
Question 7 · Straight lines · A10
State the y-intercept of y = −3x + 7.
Answer:
At the y-axis x = 0.
y = −3 × 0 + 7 = 7, so the intercept is (0, 7).
Question 8 · Straight lines · A9
Find the equation of a horizontal line through (2, 6).
Answer:
Every point on a horizontal line has the same y-coordinate.
Through (2, 6), the equation is y = 6.
Question 9 · Straight lines · A9
Find the equation of a vertical line through (−4, 3).
Answer:
Every point on a vertical line has the same x-coordinate.
Through (−4, 3), the equation is x = −4.
Question 10 · Straight lines · A9
Find the gradient of the line through (1, 2) and (3, 8).
Answer:
Gradient = change in y ÷ change in x.
(8 − 2) ÷ (3 − 1) = 6 ÷ 2 = 3.
Question 11 · Straight lines · A9
Find the equation of the line with gradient 2 through (0, −1).
Answer:
Use y = mx + c with m = 2.
The y-intercept is −1, so y = 2x − 1.
Question 12 · Straight lines · A9
Find the equation of the line through (1, 3) with gradient 4.
Answer:
Use y = 4x + c and substitute (1, 3).
3 = 4 + c, so c = −1.
The equation is y = 4x − 1.
Question 13 · Straight lines · A9
Find the equation of the line through (0, 5) and (2, 9).
Answer:
Gradient = (9 − 5) ÷ (2 − 0) = 2.
The y-intercept is 5, so y = 2x + 5.
Question 14 · Straight lines · A9
Which line is parallel to y = 3x + 2: y = −3x + 1 or y = 3x − 5?
Answer:
Parallel lines have equal gradients.
y = 3x − 5 and y = 3x + 2 both have gradient 3.
Question 15 · Straight lines · A9 · Higher only
Find the gradient of a line perpendicular to y = 2x + 7.
Answer:
Perpendicular gradients multiply to −1.
The negative reciprocal of 2 is −12.
Question 16 · Straight lines · A9 · Higher only
Are y = 4x − 3 and y = −x4 + 6 perpendicular?
Answer:
The gradients are 4 and −14.
4 × (−14) = −1, so the lines are perpendicular.
Question 17 · Straight lines · A10
Where does y = 2x − 6 cross the x-axis?
Answer:
At the x-axis y = 0.
0 = 2x − 6 gives 2x = 6 and x = 3.
The crossing is (3, 0).
Question 18 · Straight lines · A10
Find the gradient of the line from (2, 7) to (6, −1).
Answer:
Gradient = change in y ÷ change in x.
(−1 − 7) ÷ (6 − 2) = −8 ÷ 4 = −2.
Question 19 · Straight lines · A9
Does (3, 11) lie on y = 4x − 1?
Answer:
Substitute x = 3: y = 4 × 3 − 1 = 11.
This matches the point's y-coordinate, so yes.
Question 20 · Straight lines · A9
Find where y = x + 2 and y = 2x − 1 intersect.
Answer:
At an intersection the y-values are equal: x + 2 = 2x − 1.
Solve x = 3; then y = 3 + 2 = 5.
Intersection: (3, 5).
Question 21 · Quadratics · A11
Find the roots of y = (x − 2)(x + 4).
Answer:
At a root y = 0, so (x − 2)(x + 4) = 0.
Either x − 2 = 0 or x + 4 = 0.
The roots are x = 2 and x = −4.
Question 22 · Quadratics · A11
What is the y-intercept of y = x² − 5x + 6?
Answer:
At the y-axis x = 0.
y = 0² − 5 × 0 + 6 = 6, so the intercept is (0, 6).
Question 23 · Quadratics · A11
State the axis of symmetry of a quadratic with roots 1 and 7.
Answer:
The axis lies halfway between the roots.
(1 + 7) ÷ 2 = 4, so x = 4.
Question 24 · Quadratics · A11 · Higher only
Give the turning point of y = (x − 3)² + 2.
Answer:
A square is at least zero and is zero when x = 3.
The minimum y-value is 2, at (3, 2).
Question 25 · Quadratics · A11 · Higher only
Find the turning point of y = x² − 4x + 7.
Answer:
Complete the square: x² − 4x + 7 = (x − 2)² + 3.
The minimum occurs when x = 2, giving (2, 3).
Question 26 · Quadratics · A11
How many real roots does y = x² + 1 have?
Answer:
For every real x, x² ≥ 0, so x² + 1 ≥ 1.
The graph never reaches y = 0 and has no real roots.
Question 27 · Quadratics · A12
For y = x², find the y-values when x = −2 and x = 3.
Answer:
Substitute x = −2: y = (−2)² = 4.
Substitute x = 3: y = 3² = 9.
Question 28 · Quadratics · A12
Does y = −x² have a maximum or a minimum?
Answer:
Since x² ≥ 0, −x² ≤ 0.
The largest value is 0, achieved at (0, 0).
Question 29 · Other graphs · A12
If y = 1x, find y when x = −4.
Answer:
Substitute x = −4: y = 1−4.
Therefore y = −14.
Question 30 · Other graphs · A12
For y = x³, find y when x = −2.
Answer:
Substitute x = −2: y = (−2)³.
(−2) × (−2) × (−2) = −8.
Question 31 · Other graphs · A12 · Higher only
If y = 2ˣ, find y when x = −2.
Answer:
Substitute x = −2: y = 2⁻².
2⁻² = 12² = 14.
Question 32 · Other graphs · A12 · Higher only
What is the period of y = sin x when x is measured in degrees?
Answer:
The sine graph repeats after one full turn.
One full turn is 360°, so the period is 360°.
Question 33 · Other graphs · A12 · Higher only
What is cos 180°?
Answer:
At 180° the corresponding point on the unit circle has x-coordinate −1.
Therefore cos 180° = −1.
Question 34 · Graph changes · A13 · Higher only
A point (2, 5) is on y = f(x). What corresponding point is on y = f(x) + 3?
Answer:
Adding 3 to f(x) shifts every point 3 units up.
Keep x = 2 and add 3 to y: (2, 8).
Question 35 · Graph changes · A13 · Higher only
A point (2, 5) is on y = f(x). What corresponding point is on y = f(x − 4)?
Answer:
Replacing x by x − 4 shifts the graph 4 units right.
Add 4 to the x-coordinate: (6, 5).
Question 36 · Graph changes · A13 · Higher only
A point (−1, 7) is on y = f(x). What corresponding point is on y = −f(x)?
Answer:
Multiplying f(x) by −1 reflects the graph in the x-axis.
Keep x and change the sign of y: (−1, −7).
Question 37 · Graph changes · A13 · Higher only
A point (3, −2) is on y = f(x). What corresponding point is on y = f(−x)?
Answer:
Replacing x by −x reflects the graph in the y-axis.
Change the sign of x and keep y: (−3, −2).
Question 38 · Real graphs · A14
A distance-time graph rises by 60 km over 2 hours at constant speed. What is the speed?
Answer:
Speed is the gradient: change in distance ÷ change in time.
60 ÷ 2 = 30 km/h.
Question 39 · Real graphs · A14
A horizontal section on a distance-from-home versus time graph means what?
Answer:
A horizontal section has zero change in distance from home.
The distance from home remains unchanged during that interval.
Question 40 · Real graphs · A14
A taxi fare is £3 plus £2 per mile. Write the fare C for m miles.
Answer:
The fixed part is £3 and the mileage part is £2 × m.
Adding them gives C = 2m + 3.
Question 41 · Real graphs · A14
The points (0, 0) and (4, 20) lie on a straight distance-time graph. What is its gradient?
Answer:
Gradient = change in distance ÷ change in time.
(20 − 0) ÷ (4 − 0) = 5 distance units per time unit.
Question 42 · Real graphs · A15 · Higher only
A velocity-time graph is a rectangle of height 8 m/s and width 5 s. Find the distance travelled.
Answer:
Distance is the area under the non-negative velocity-time graph.
Rectangle area = 8 × 5 = 40 m.
Question 43 · Real graphs · A15 · Higher only
Velocity increases uniformly from 0 to 12 m/s in 4 s. Find the distance travelled.
Answer:
Distance is the area under the velocity-time graph.
Triangle area = 12 × 4 × 12 = 24 m.
Question 44 · Real graphs · A15 · Higher only
Velocity changes uniformly from 2 m/s to 8 m/s in 3 s. Find the acceleration.
Answer:
Acceleration = change in velocity ÷ time.
(8 − 2) ÷ 3 = 2 m/s².
Question 45 · Circles · A16 · Higher only
What is the radius of the circle x² + y² = 25?
Answer:
A circle centred at the origin has equation x² + y² = r².
r² = 25, so the positive radius is r = 5.
Question 46 · Circles · A16 · Higher only
Does (3, 4) lie on the circle x² + y² = 25?
Answer:
Substitute the coordinates: 3² + 4² = 9 + 16 = 25.
The equation is satisfied, so (3, 4) lies on the circle.
Question 47 · Circles · A16 · Higher only
State the centre of x² + y² = 16.
Answer:
The equation x² + y² = r² represents a circle centred at the origin.
Its centre is (0, 0).
Question 48 · Circles · A16 · Higher only
Find the gradient of the tangent to x² + y² = 25 at (3, 4).
Answer:
The radius from (0, 0) to (3, 4) has gradient 43.
The tangent is perpendicular, so its gradient is −34.
Question 49 · Interpretation · A10
A line has gradient −2 and y-intercept 5. Give its equation.
Answer:
Use y = mx + c, with gradient m = −2 and intercept c = 5.
The equation is y = −2x + 5.
Question 50 · Interpretation · A10
What does a negative gradient mean as x increases?
Answer:
A negative gradient means change in y is negative when change in x is positive.
As x increases, y decreases.

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