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.
Gradients and intercepts of linear functions worksheet — GCSE Foundation
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- 1.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.y = 3x + 6
- 2.Work out the coordinates of the point where the graph of y = 2x − 6 crosses the x-axis.y = 2x − 6
- 3.A car's fuel tank contains 50 litres when it starts a journey. After travelling for 2 hours, it contains 38 litres, and fuel is used at a constant rate. Work out the rate at which fuel is used, in litres per hour.
- 4.A water tank empties according to y = −5x + 200, where y is the number of litres left after x minutes. Work out the coordinates of the point where this graph crosses the x-axis.y = -5x + 200
- 5.A candle is 30 cm tall when lit and burns at a constant rate. After burning for 5 minutes, its height is 20 cm. The height h cm after t minutes is given by h = 30 − mt. Work out the value of m.
- 6.A straight line passes through the points (0, 4) and (1, 9). Write down the gradient of the line.
- 7.Work out the gradient of the straight line with equation y = (2x/5) − 3.
- 8.A straight line passes through the points (2, 7) and (5, 16). Work out the value of y when x = 0.
- 9.A tank already contains some water and is then filled at a constant rate. After 2 minutes it contains 50 litres in total; after 7 minutes it contains 150 litres in total. Work out the rate at which water is added, in litres per minute.
- 10.Write down the gradient of the line with equation y = −x + 3.y = −x + 3
- 11.A car passes the 15 km marker on a road and passes the 195 km marker 3 hours later. Work out the gradient of the distance-time graph, that is the rate of change of distance with time.
- 12.Work out the gradient of the straight line with equation 3y = 12 − 6x.
- 13.A plumber charges a call-out fee plus an hourly rate for a job. The total cost, £C, for a job lasting h hours is given by the formula C = 40 + 25h. What does the number 40 represent in this formula?
- 14.Write down the coordinates of the point where the line y = 2x + 4 crosses the y-axis.y = 2x + 4
- 15.A mobile phone plan charges a fixed monthly fee plus an amount for each gigabyte of data used. The total cost £C for using g gigabytes in a month is given by C = 15 + 2g. What does the number 2 represent in this formula?
Answer key
- (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) (3, 0) — Method: every point on the x-axis has y-coordinate 0, so substituting y = 0 into the equation and solving gives the x-coordinate of the crossing point. Working: 0 = 2x − 6 gives 2x = 6, so x = 6 ÷ 2 = 3 and the graph crosses the x-axis at (3, 0). Answer: (3, 0). The distractors: (0, −6) is where the graph crosses the y-axis, found by substituting x = 0 rather than y = 0; (−3, 0) comes from moving the 6 across the equals sign without changing its sign, giving 2x = −6; (6, 0) comes from reading the constant straight off as the crossing point and never dividing by the gradient 2.
- (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.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (a) 2 — The candle's height falls from 30 cm to 20 cm, a drop of 10 cm, over 5 minutes, so m = 10 ÷ 5 = 2. 10 comes from using the drop in height but forgetting to divide by the time. 0.5 comes from dividing the time by the drop instead of the drop by the time (5 ÷ 10). 4 comes from dividing the final height by the time, 20 ÷ 5, instead of using the drop in height.
- (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.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (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.
- (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.
- (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.
- (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) 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.
- (d) (0, 4) — Method: in the form y = mx + c the constant c is the y-coordinate of the point where the line crosses the y-axis, and every point on that axis has x-coordinate 0. Working: comparing y = 2x + 4 with y = mx + c gives c = 4, and substituting x = 0 confirms y = 2 × 0 + 4 = 4, so the crossing point is (0, 4). Answer: (0, 4). The distractors: (4, 0) is the same pair of numbers written in the wrong order and is a point on the x-axis; (0, 2) comes from reading the gradient 2 as the intercept, confusing m with c; (−2, 0) is where the line crosses the x-axis, found by solving 0 = 2x + 4 instead.
- (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.
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