Printable · GCSE Foundation · ages 14-16
Gradients and intercepts of linear functions worksheet — GCSE Foundation
Fifteen questions on "gradients and intercepts of linear functions" — DfE statement A10. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Gradients and intercepts of linear functions worksheet — GCSE Foundation
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- (b) 3 — Method: rearrange the equation into the form y = mx + c first, then read off the gradient. Working: dividing 4y = 12x + 20 by 4 gives y = 3x + 5, so the gradient is 3. The value 12 comes from reading the coefficient of x before dividing the equation by 4. The value 20 comes from using the constant term of the unsimplified equation instead of the gradient. The value 5 comes from finding the y-intercept, 20 / 4 = 5, and giving that instead of the gradient.
- (a) 0.25 — Method: gradient = vertical rise ÷ horizontal distance. Working: gradient = 3 ÷ 12 = 0.25. Answer: the gradient of the ramp is 0.25. 4 comes from inverting the gradient, dividing the horizontal distance by the vertical rise instead of the other way round. 9 comes from subtracting the rise from the horizontal distance instead of dividing. 36 comes from multiplying the rise and the horizontal distance instead of dividing.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
- (b) −1 — Method: in the form y = mx + c the gradient m is the number multiplying x, and an x term written with no number in front of it has a coefficient of 1. Working: y = −x + 3 is the same equation as y = (−1)x + 3, so comparing it with y = mx + c gives m = −1 and c = 3. Answer: −1. The distractors: 1 comes from taking the coefficient as 1 and leaving the minus sign out of the answer; 3 comes from reading the constant as the gradient, confusing m with c; −3 comes from moving the minus sign across to the constant term and then reading the gradient off that term instead of the x term.
- (d) −5 — In y = mx + c, the y-intercept is c, the constant term, not the coefficient of x. Here c = −5, so the y-intercept is −5. 4 is Beth's value, the coefficient of x (the gradient), not the intercept. 5 comes from taking the constant term but dropping its negative sign. −4 comes from taking the gradient but giving it the sign of the intercept.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (d) The fixed call-out fee, charged before the hourly rate. — Method: compare the formula with C = (fixed charge) + (rate) × h, where the fixed charge is the part that does not depend on h. Working: in C = 40 + 25h, the term 25h depends on the number of hours, h, but 40 does not change however many hours the job takes. Answer: 40 is the fixed call-out fee. 'The hourly rate' confuses the fixed term with the coefficient of h, which is actually 25. 'The total cost for a job lasting 1 hour' comes from substituting h = 1 into the formula (40 + 25 = 65) rather than reading off the constant term. 'The number of hours worked before charging starts' wrongly treats the constant, which is in pounds, as if it were measured in hours.
- (c) 5 — Method: gradient = change in y ÷ change in x between the two points. Working: change in y = 9 − 4 = 5, change in x = 1 − 0 = 1, so gradient = 5 ÷ 1 = 5. Answer: the gradient is 5. 13 comes from adding the y-values instead of subtracting them. 4 comes from using the y-value at x = 0 instead of the change in y between the two points. 1/5 comes from inverting the gradient, dividing the change in x by the change in y instead of the other way round.
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (c) The cost, in pounds, for each extra gigabyte of data used — In C = 15 + 2g, the number multiplying g is the gradient, which gives the extra cost for each extra unit of g — here, £2 for each extra gigabyte. 'The fixed monthly fee, in pounds' describes the constant term 15, not the coefficient of g. 'The total number of gigabytes included in the plan' misreads the coefficient as a quantity of data rather than a cost per gigabyte. 'The cost … for each extra 2 gigabytes' doubles the unit the coefficient actually applies to — it is the cost for each single extra gigabyte.
- (c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
- (d) 4 — Method: compare the equation with the general form y = mx + c after writing the x term first, then read off the coefficient of x. Working: y = 6 + 4x can be written as y = 4x + 6, so comparing with y = mx + c gives m = 4. Answer: the gradient is 4. The value 6 comes from reading off the constant, written first in the equation, and treating it as the gradient because it comes before the x term. The value 10 comes from adding the two numbers in the equation, 6 + 4, instead of reading off the coefficient of x. The value 24 comes from multiplying the two numbers in the equation, 6 × 4, instead of reading off the coefficient of x.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
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