OCR GCSE Maths · OCR 7

Graphs of Equations and Functions

Maths questions with worked answers with answers for OCR 7 Graphs of Equations and Functions in OCR GCSE Maths (spec J560). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

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Question 1 · Coordinates · OCR 7.01a
Find the midpoint of (2, 7) and (8, −1).
Answer:
Average the x-values: 2 + 82 = 5. Average the y-values: 7 + (−1)2 = 3.
Midpoint (5, 3).
Question 2 · Coordinates · OCR 7.01a
The midpoint of AB is (4, 6). A is (1, 2). Find B.
Answer:
From A to the midpoint: +3 in x and +4 in y.
B = (4 + 3, 6 + 4) = (7, 10).
Question 3 · Coordinates · OCR 7.01a
Which quadrant contains the point (−3, 5)?
Answer:
Negative x, positive y.
The second quadrant (top left).
Question 4 · Coordinates · OCR 7.01a
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 5 · Coordinates · OCR 7.01a
What is the x-coordinate of (7, −2)?
Answer:
Coordinates are written in the order (x, y).
The first coordinate is 7.
Question 6 · Coordinates · OCR 7.01a
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 7 · Graph shapes · OCR 7.01c
What shape is the graph of y = x³?
Answer:
A cubic curve.
It rises from bottom left, flattens through the origin and rises to the top right.
Question 8 · Graph shapes · OCR 7.01c
Describe the graph of y = 1x.
Answer:
A reciprocal graph.
Two separate curves in opposite quadrants; it never touches either axis.
Question 9 · Graph shapes · OCR 7.01c · Higher only
Sketch the key features of y = 2ˣ.
Answer:
It passes through (0, 1).
It rises steeply for positive x and approaches the x-axis for negative x without touching it.
Question 10 · Trig graphs · OCR 7.01c · Higher only
State the maximum value and period of y = sin x.
Answer:
sin x varies between −1 and 1.
Maximum 1; period 360°.
Question 11 · Trig graphs · OCR 7.01c · Higher only
For 0° ≤ x ≤ 360°, where does y = cos x cross the x-axis?
Answer:
cos x = 0.
At x = 90° and x = 270°.
Question 12 · Trig graphs · OCR 7.01c · Higher only
Given sin 30° = 0.5, find another angle between 0° and 360° with sine 0.5.
Answer:
Use the symmetry of the sine graph: 180° − 30°.
150°.
Question 13 · Graph shapes · OCR 7.01c
Which graph is a straight line: y = 3x + 1, y = x² or y = 1x?
Answer:
Only linear equations give straight lines.
y = 3x + 1.
Question 14 · Quadratics · OCR 7.01c
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 15 · Quadratics · OCR 7.01c
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 16 · Circles · OCR 7.01f · Higher only
Write the equation of the circle with centre (0, 0) and radius 6.
Answer:
x² + y² = r².
x² + y² = 36.
Question 17 · Circles · OCR 7.01f · Higher only
Find the radius of x² + y² = 50.
Answer:
r² = 50.
r = √50 = 5√2.
Question 18 · Circles · OCR 7.01f · Higher only
Show that (3, 4) lies on x² + y² = 25.
Answer:
3² + 4² = 9 + 16.
= 25, so the point lies on the circle.
Question 19 · Circles · OCR 7.01f · Higher only
Find the gradient of the tangent to x² + y² = 25 at (3, 4).
Answer:
Gradient of radius = 43. The tangent is perpendicular.
Gradient = −34.
Question 20 · Circles · OCR 7.01f · 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 21 · Gradient · OCR 7.02a
Find the gradient of the line through (1, 3) and (5, 11).
Answer:
Gradient = change in ychange in x = 11 − 35 − 1.
84 = 2.
Question 22 · Straight lines · OCR 7.02a
Write the gradient and y-intercept of y = 5x − 3.
Answer:
Compare with y = mx + c.
Gradient 5, y-intercept (0, −3).
Question 23 · Straight lines · OCR 7.02a
Find the gradient and y-intercept of 2y + 6x = 8.
Answer:
Rearrange: 2y = −6x + 8, so y = −3x + 4.
Gradient −3, y-intercept (0, 4).
Question 24 · Equation of a line · OCR 7.02a
Find the equation of the line with gradient 4 through (2, 5).
Answer:
y = 4x + c. Substitute: 5 = 8 + c, so c = −3.
y = 4x − 3.
Question 25 · Graphs · OCR 7.02a
Where does the line y = 3x − 6 cross the x-axis?
Answer:
On the x-axis, y = 0: 3x − 6 = 0.
x = 2, so (2, 0).
Question 26 · Roots and intercepts · OCR 7.02a
Write the coordinates of where y = x² − 9 crosses the axes.
Answer:
y-axis: x = 0 gives y = −9. x-axis: x² = 9.
(0, −9), (3, 0) and (−3, 0).
Question 27 · Solving graphically · OCR 7.02a · Higher only
How can you use a graph to solve x² − 2x − 3 = 0?
Answer:
Plot y = x² − 2x − 3.
Read off the x-values where the curve crosses the x-axis (x = −1 and x = 3).
Question 28 · Solving graphically · OCR 7.02a
How do you use graphs to solve a pair of simultaneous equations?
Answer:
Draw both lines on the same axes.
The solution is the point where they intersect.
Question 29 · Parallel lines · OCR 7.02b
Write the equation of a line parallel to y = 2x + 1 passing through (0, −4).
Answer:
Parallel lines have the same gradient, 2.
y = 2x − 4.
Question 30 · Perpendicular lines · OCR 7.02b · Higher only
Find the gradient of a line perpendicular to y = 23x + 1.
Answer:
Perpendicular gradients multiply to −1.
−32.
Question 31 · Perpendicular lines · OCR 7.02b · Higher only
Find the equation of the line perpendicular to y = 4x − 2 through (8, 1).
Answer:
Gradient = −14. 1 = −14 × 8 + c, so 1 = −2 + c, c = 3.
y = −14x + 3.
Question 32 · Straight lines · OCR 7.02b · 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 33 · Transformations · OCR 7.03a · Higher only
Describe the transformation that maps y = f(x) to y = f(x) + 3.
Answer:
3 is added to every y-value.
Translation by 03.
Question 34 · Transformations · OCR 7.03a · Higher only
y = x² has turning point (0, 0). State the turning point of y = (x + 5)² − 1.
Answer:
Translation by −5−1.
(−5, −1).
Question 35 · Graph changes · OCR 7.03a · 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 36 · Real-life graphs · OCR 7.04a
On a distance–time graph, what does the gradient represent?
Answer:
Distance ÷ time.
Speed.
Question 37 · Real-life graphs · OCR 7.04a
On a distance–time graph, what does a horizontal line show?
Answer:
Distance is not changing.
The object is stationary.
Question 38 · Real-life graphs · OCR 7.04a
A taxi charges £3 plus £2 per mile. Write a formula for the cost C of m miles.
Answer:
Fixed charge £3, plus 2 for each mile.
C = 2m + 3.
Question 39 · Real-life graphs · OCR 7.04a
In the formula C = 2m + 3, what does the 3 represent on the graph?
Answer:
It is the y-intercept.
The fixed charge of £3 before any miles are travelled.
Question 40 · Real-life graphs · OCR 7.04a
A car travels 120 km in the first 2 hours. Find its speed from the graph's gradient.
Answer:
Gradient = 1202.
60 km/h.
Question 41 · Gradient · OCR 7.04a
The gradient of a distance–time graph is 15. What does this mean?
Answer:
Gradient = distance ÷ time.
A speed of 15 (e.g. 15 m/s).
Question 42 · Rates of change · OCR 7.04a · Higher only
What does the gradient of a tangent to a curve show?
Answer:
The rate of change at that instant.
For example, the speed at a moment on a distance–time curve.
Question 43 · Real graphs · OCR 7.04a
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 44 · Velocity–time graphs · OCR 7.04b · Higher only
A car accelerates uniformly from 0 to 20 m/s in 8 seconds. Find the distance travelled.
Answer:
The area under the graph is a triangle.
12 × 8 × 20 = 80 m.
Question 45 · Rates of change · OCR 7.04b · Higher only
How do you estimate the gradient of a curve at a point?
Answer:
Draw a tangent to the curve at that point.
Find the gradient of the tangent: change in y ÷ change in x.
Question 46 · Area under a curve · OCR 7.04b · Higher only
How can you estimate the area under a curve?
Answer:
Split the area into trapezia (or strips).
Add the areas of the shapes.
Question 47 · Rates of change · OCR 7.04b · Higher only
A water level rises from 12 cm to 30 cm in 6 minutes. Find the average rate of change.
Answer:
Change = 18 cm.
18 ÷ 6 = 3 cm per minute.
Question 48 · Real graphs · OCR 7.04b · 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 49 · Real graphs · OCR 7.04b · 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 50 · Real graphs · OCR 7.04b · 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.

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