Applications: Graphs of Equations

Each gym's cost is a straight line: the monthly charge is the y-intercept and the charge for each visit is the gradient. The two gyms cost the same at the point where the lines cross, because that point satisfies both equations at once. On either side of the crossing, the lower line is the cheaper gym.

1. Let x be the number of visits in a month and y the cost in dollars. Gym A costs y = 4x + 10 and Gym B costs y = 2x + 30.
2. The two gyms cost the same where the lines cross. At that point both equations hold, so the two expressions for y are equal: 4x + 10 = 2x + 30.
3. Subtract 2x and then 10 from both sides: 2x = 20, so x = 10, and y = 4 × 10 + 10 = 50. (a) The gyms cost the same for 10 visits, when each costs $50. Check in Gym B: 2 × 10 + 30 = 50.
4. For 14 visits, read both lines at x = 14. Gym A costs 4 × 14 + 10 = 66 dollars and Gym B costs 2 × 14 + 30 = 58 dollars.
5. (b) To the right of the crossing the line of Gym B is below the line of Gym A, so Gym B is cheaper for Dinesh, by 66 − 58 = $8.

answer(a) 10 visits, when each gym costs $50; (b) Gym B, by $8

techniqueIntersections of Graphs · Simultaneous Equations · y = mx + c

examsSAT · GCSE Higher · H2

Two points fix a straight line. The gradient is the rise divided by the run between the two points, and the point-slope form turns the gradient and one point into the equation. In y = mx + c the gradient m is the charge for each kilometer, and the intercept c is the fare when x = 0, which is the charge for starting the journey.

1. Find the gradient from the two points. From (2, 7) to (6, 13) the line rises 13 − 7 = 6 while it runs 6 − 2 = 4, so m = 64 = 1.5.
2. Use the point-slope form with the point (2, 7): y − 7 = 1.5(x − 2). Expand the bracket: y − 7 = 1.5x − 3, so y = 1.5x + 4.
3. (a) The equation is y = 1.5x + 4. The gradient 1.5 means that each kilometer adds $1.50 to the fare. The y-intercept 4 means that every journey starts with a charge of $4. Check with the other point: 1.5 × 6 + 4 = 13.
4. For a fare of $19, put y = 19 into the equation: 19 = 1.5x + 4. Subtract 4 from both sides: 1.5x = 15. Divide both sides by 1.5: x = 10.
5. (b) The journey is 10 km long. Check: 1.5 × 10 + 4 = 19, so the point (10, 19) is on the line.

answer(a) y = 1.5x + 4: each kilometer costs $1.50 and every journey starts at $4; (b) 10 km

techniquey = mx + c · Point-Slope Form of a Line · A Line in Graph, Table and Equation

examsO-Level · SAT · GCSE Higher

The total cost is a sum of two terms, one for each kind of food, so the equation has the standard form Ax + By = C. In this form each intercept comes from putting the other variable equal to zero. An intercept is an order of only one kind of food, and every other point on the line is a mixed order that spends the same amount.

1. x pizzas cost 8x dollars and y salads cost 6y dollars. All of the money is spent, so 8x + 6y = 120.
2. Put x = 0 to find where the line meets the y-axis: 6y = 120, so y = 20. Put y = 0 to find where it meets the x-axis: 8x = 120, so x = 15.
3. (a) The equation is 8x + 6y = 120, and its graph joins (0, 20) and (15, 0). The point (0, 20) means 20 salads and no pizzas. The point (15, 0) means 15 pizzas and no salads.
4. For 9 pizzas, put x = 9 into the equation: 72 + 6y = 120. Subtract 72 from both sides: 6y = 48, so y = 8.
5. (b) The committee can buy 8 salads. Check: 8 × 9 + 6 × 8 = 72 + 48 = 120, so the point (9, 8) is on the line.

answer(a) 8x + 6y = 120, which meets the axes at (15, 0) and (0, 20): 15 pizzas and no salads, or 20 salads and no pizzas; (b) 8 salads

techniqueWriting a Line as Ax + By = C · The Coordinate Plane

examsSAT

On a distance-time graph the gradient of a straight part is the speed on that part, because it is the distance covered divided by the time taken. A level part has a gradient of zero, which is a rest. The average speed for the whole outing is not the mean of the two speeds: it is the total distance divided by the total time.

1. The graph has three straight parts. It rises from (0, 0) to (2, 36), stays level until (3, 36), and falls to (4.5, 0). The gradient of each part is the speed on that part.
2. On the way out Mei covers 36 km in 2 hours, so the gradient is 362 = 18. Her speed is 18 km/h.
3. From 2 hours to 3 hours the graph is level, so she rests for 1 hour. On the way back she covers 36 km in 4.5 − 3 = 1.5 hours, so her speed is 361.5 = 24 km/h. The gradient of this part is −24, which is negative because her distance from home is decreasing. (a) She rides at 18 km/h on the way out and at 24 km/h on the way back.
4. For the whole outing, the total distance is 36 + 36 = 72 km, and the total time, including the rest, is 4.5 hours.
5. (b) The average speed is 724.5 = 16 km/h. Check: 16 × 4.5 = 72.

answer(a) 18 km/h on the way out and 24 km/h on the way back; (b) 16 km/h

techniqueDistance-Time Graphs

examsSAT · GCSE Higher

On a speed-time graph the gradient is the change in speed in each second, which is the acceleration. The distance traveled is the area between the graph and the time axis, because a speed multiplied by a time is a distance. The area is found by splitting it into a triangle, a rectangle and a triangle.

1. The acceleration is the gradient of the first part of the graph. The speed rises by 20 m/s in 10 seconds, so the acceleration is 2010 = 2 m/s². (a) The acceleration is 2 m/s².
2. The distance traveled is the area under the graph. Vertical lines at 10 seconds and at 40 seconds split this area into a triangle, a rectangle and another triangle.
3. The first triangle has area 12 × 10 × 20 = 100. The rectangle is 40 − 10 = 30 seconds wide, so its area is 30 × 20 = 600. The second triangle is 20 seconds wide, so its area is 12 × 20 × 20 = 200.
4. (b) The distance between the stations is 100 + 600 + 200 = 900 m. Check: the whole shape is a trapezium with parallel sides of 60 and 30 and a height of 20, and 12 × (60 + 30) × 20 = 900.

answer(a) 2 m/s²; (b) 900 m

techniqueSpeed-Time Graphs

examsSAT · GCSE Higher

The job is a fixed amount of work, 24 worker-hours. The number of workers multiplied by the number of hours always gives this amount, so the two quantities are inversely proportional and the graph is the reciprocal curve y = 24x. Reading the curve one way gives the time for a team, and reading it the other way gives the team for a time.

1. One worker takes 24 hours, so the job is 24 worker-hours of work. With x workers for y hours, xy = 24, so y = 24x.
2. Plot a few points: 1 worker takes 24 hours, 2 workers take 12 hours and 4 workers take 6 hours. Each time the number of workers doubles, the time halves, so the graph is a curve and not a straight line.
3. (a) y = 24x. For 6 workers, y = 246 = 4, so 6 workers take 4 hours.
4. For a time of 3 hours, draw the line y = 3 across to the curve. At the point where they meet, 3 = 24x. Multiply both sides by x: 3x = 24, so x = 8.
5. (b) 8 workers are needed. Check: 8 × 3 = 24 worker-hours. More workers take less time still, but 24x is never 0, so the curve never meets the x-axis.

answer(a) y = 24x, and 6 workers take 4 hours; (b) 8 workers

techniqueReciprocal Graphs

examsO-Level · SAT · GCSE Higher · H2

The graph of y = 5x² is a curve, so the ball does not fall at one steady speed. Over an interval, the average speed is the distance fallen divided by the time taken. On the graph this is the gradient of the chord, the straight line that joins the two points at the ends of the interval.

1. Find the points on the curve at the ends of the first interval. At x = 1, y = 5 × 1² = 5. At x = 3, y = 5 × 3² = 45.
2. The average speed is the gradient of the chord from (1, 5) to (3, 45): 45 − 53 − 1 = 402 = 20. (a) The average speed between x = 1 and x = 3 is 20 m/s.
3. At x = 4, y = 5 × 4² = 80. The chord from (3, 45) to (4, 80) has gradient 80 − 454 − 3 = 351 = 35.
4. (b) The average speed between x = 3 and x = 4 is 35 m/s. The second chord is steeper than the first, which shows that the ball falls faster and faster.

answer(a) 20 m/s; (b) 35 m/s, so the ball is falling faster and faster

techniqueThe Average Rate of Change Over an Interval · The Graphs of y = axⁿ · Quadratic and Cubic Graphs

examsSAT

On a curve the rate of change is different at every point. The rate at one moment is the gradient of the curve at that point, and the gradient of the curve is the gradient of the tangent drawn there. The average rate over an interval is the gradient of the chord across the interval. Comparing the two shows how the cooling slows down.

1. The rate of cooling at 10 minutes is the gradient of the curve at (10, 50). The tangent touches the curve at that point and has the same gradient there, so the gradient is read from the tangent.
2. Choose two points on the tangent that are far apart, (0, 60) and (30, 30). The gradient is 30 − 6030 − 0 = −3030 = −1.
3. (a) The gradient is negative because the temperature is falling. After 10 minutes the tea is cooling at about 1^°C per minute.
4. The average rate over the first 10 minutes is the gradient of the chord from (0, 65) to (10, 50): 50 − 6510 − 0 = −1510 = −1.5.
5. (b) Over the first 10 minutes the tea cools by 1.5^°C per minute on average. The curve is steepest at the start and becomes less steep, so the tea cools fastest when it is hottest, and the chord over the first 10 minutes is steeper than the tangent at x = 10.

answer(a) About 1^°C per minute; (b) 1.5^°C per minute, which is greater because the curve is steeper at the start than at x = 10

techniqueEstimating a Gradient with a Drawn Tangent · The Average Rate of Change Over an Interval

examsO-Level

A distance is never negative, whichever side of the depot the place is on, so the distance between marker x and marker 12 is the modulus |x − 12|. Its graph is a V with its lowest point at the depot. A horizontal line at height 5 meets the V at the two places exactly 5 km away, and the part of the V on or below that line shows every place within 5 km.

1. The distance between marker x and the depot at marker 12 is the difference between the two numbers, taken as positive: y = |x − 12|. The graph is a V with its lowest point at (12, 0).
2. To the left of the depot, where x < 12, the graph is the line y = 12 − x. To the right, where x > 12, it is the line y = x − 12. A place exactly 5 km away is a point where the V meets the horizontal line y = 5.
3. On the left, 12 − x = 5 gives x = 7. On the right, x − 12 = 5 gives x = 17. (a) y = |x − 12|, and the 7 km and 17 km markers are exactly 5 km from the depot.
4. A place is within 5 km of the depot where the V is on or below the line y = 5. That is the part of the road with 7 ≤ x ≤ 17.
5. The heights of the V at the four shops are |4 − 12| = 8, |9 − 12| = 3, |15 − 12| = 3 and |18 − 12| = 6. (b) Only the shops at the 9 km and 15 km markers are within 5 km, so these two have free delivery.

answer(a) y = |x − 12|, and the 7 km and 17 km markers; (b) the shops at the 9 km and 15 km markers

techniqueThe Graph of y = |x| · Taking the Modulus of a Whole Expression

examsSAT · H2

The places exactly 10 km from the mast form a circle with its center at the origin, and by Pythagoras' theorem its equation is x² + y² = 100. A point is inside the circle when x² + y² is less than 100. The road meets the edge of the range where a point satisfies both the line and the circle, so the line is substituted into the circle.

1. A point (x, y) is √x² + y² km from the mast. On the edge of the range this distance is 10 km, so the edge is the circle x² + y² = 10², which is x² + y² = 100.
2. For the town at (5, −8), x² + y² = 5² + (−8)² = 25 + 64 = 89. (a) Yes. Since 89 < 100, the town is less than 10 km from the mast, so it is inside the circle and receives the signal.
3. Where the road meets the edge, both equations hold. Substitute y = x + 2 into the circle: x² + (x + 2)² = 100. Expand: 2x² + 4x + 4 = 100. Subtract 100 from both sides and divide both sides by 2: x² + 2x − 48 = 0.
4. Factorize: two numbers with a product of −48 and a sum of 2 are 8 and −6, so (x + 8)(x − 6) = 0, and x = −8 or x = 6. Substitute each root into y = x + 2: y = −6 when x = −8, and y = 8 when x = 6.
5. (b) The road enters the range at (−8, −6) and leaves it at (6, 8). Check in the circle: (−8)² + (−6)² = 64 + 36 = 100 and 6² + 8² = 36 + 64 = 100.

answer(a) Yes: 5² + (−8)² = 89, which is less than 100; (b) at (−8, −6) and (6, 8)

techniqueWhere a Line Meets a Circle · The Equation of a Circle

examsO-Level · GCSE Higher

The expanded equation hides the center and the radius. Completing the square in x and in y rewrites it as (x − a)² + (y − b)² = r², from which the center (a, b) and the radius r are read directly. The nearest point of a circle to a point outside it lies on the line through the center, so the shortest distance is the distance to the center minus the radius.

1. Group the x terms and the y terms, and subtract 84 from both sides: (x² − 12x) + (y² − 16y) = −84.
2. Complete each square. Half of 12 is 6, and (x − 6)² = x² − 12x + 36, so x² − 12x = (x − 6)² − 36. Half of 16 is 8, so y² − 16y = (y − 8)² − 64.
3. Substitute both: (x − 6)² − 36 + (y − 8)² − 64 = −84. Add 36 and 64 to both sides: (x − 6)² + (y − 8)² = 16.
4. (a) Since 16 = 4², the roundabout has its center at (6, 8) and a radius of 4 m.
5. The lamp post at the origin is √6² + 8² = √100 = 10 m from the center. This is more than the radius, so the lamp post is outside the roundabout, and the nearest point of the curb lies on the straight line from the lamp post to the center.
6. (b) Along that line the curb is one radius nearer than the center, so the nearest point of the curb is 10 − 4 = 6 m from the lamp post.

answer(a) Center (6, 8) and radius 4 m; (b) 6 m

techniqueCenter and Radius by Completing the Square · The Equation of a Circle

examsO-Level · SAT

In the equation x²a² + y²b² = 1 the ellipse reaches a units to each side of the center and b units above it. These give the width and the greatest height of the arch. Any other height comes from substituting the value of x and solving for y, keeping the positive root because a height is positive.

1. Compare the equation with x²a² + y²b² = 1. Here a² = 100 and b² = 36, so a = 10 and b = 6.
2. On the road y = 0, so x²100 = 1 and x = −10 or x = 10. At the center x = 0, so y²36 = 1 and y = 6. (a) The tunnel is 10 + 10 = 20 m wide at road level, and the arch is 6 m high at the center.
3. For a point 8 m from the center, put x = 8 into the equation: 64100 + y²36 = 1. Subtract 0.64 from both sides: y²36 = 0.36.
4. Multiply both sides by 36: y² = 0.36 × 36 = 12.96, so y = 3.6 or y = −3.6. A height cannot be negative, so y = −3.6 is rejected.
5. (b) The arch is 3.6 m high above a point 8 m from the center. Check: 64100 + 12.9636 = 0.64 + 0.36 = 1.

answer(a) 20 m wide and 6 m high; (b) 3.6 m

techniqueThe Equation of an Ellipse

examsH2

The architect's formula describes only the right-hand slope. The left-hand slope is its mirror image in the y-axis, so a point 4 m to the left of the ridge is exactly as high as the point 4 m to the right of it. Replacing x by |x| in the formula says this in one equation, because x and −x have the same modulus and so give the same height. A level found on the right-hand slope is then reflected to find the matching point on the left.

1. The height at x = −4 must equal the height at x = 4, and replacing x by |x| does exactly this, since |−4| = |4| = 4. The whole roof line is y = (|x| − 8)²16 for −8 ≤ x ≤ 8.
2. At the ridge x = 0, so y = (0 − 8)²16 = 6416 = 4: the ridge is 4 m above the eaves. At each eave |x| = 8, so y = 0.
3. At x = −4, |x| = 4, so y = (4 − 8)²16 = 1616 = 1. (a) The roof line is y = (|x| − 8)²16 for −8 ≤ x ≤ 8, and the roof is 1 m above the eaves at x = −4.
4. The beam meets the right-hand slope where (x − 8)²16 = 2.25. Multiply both sides by 16: (x − 8)² = 36, so x − 8 = 6 or x − 8 = −6, which gives x = 14 or x = 2.
5. The right-hand slope runs only from x = 0 to x = 8, so x = 14 is rejected: it lies beyond the eaves, where there is no roof. The beam meets the right-hand slope at x = 2, and its mirror image puts the other end on the left-hand slope at x = −2.
6. (b) The beam runs from x = −2 to x = 2, so it is 2 − (−2) = 4 m long. Check: at x = −2, y = (2 − 8)²16 = 3616 = 2.25.

answer(a) y = (|x| − 8)²16 for −8 ≤ x ≤ 8, and 1 m; (b) 4 m

techniqueReplacing x with a Modulus · The Graph of y = |x|

examsSAT · H2

A straight line that touches a circle at one point is a tangent, and a tangent is perpendicular to the radius drawn to the point where it touches. So the gradient of the path comes from the gradient of the radius from the center to P: the two gradients multiply to −1. With that gradient and the point P, the point-slope form gives the equation of the path, and each fence is met where one of the coordinates is 0.

1. Compare the equation with (x − a)² + (y − b)² = r²: the center of the pond is C(7, 7) and the radius is √25 = 5 m. The point P(4, 3) is on the edge, since (4 − 7)² + (3 − 7)² = 9 + 16 = 25.
2. The radius from P(4, 3) to C(7, 7) has gradient 7 − 37 − 4 = 43.
3. The path touches the pond at P, so it is perpendicular to the radius there. Perpendicular gradients multiply to −1, so the gradient of the path is −34. Check: 43 × (−34) = −1.
4. Use the point-slope form through P(4, 3): y − 3 = −34(x − 4). Multiply both sides by 4: 4y − 12 = −3x + 12. (a) The path is the line 3x + 4y = 24.
5. The south fence is the x-axis, where y = 0: 3x = 24, so x = 8. The west fence is the y-axis, where x = 0: 4y = 24, so y = 6. The path meets the fences at (8, 0) and (0, 6).
6. (b) The path runs from (0, 6) on the west fence to (8, 0) on the south fence, and by Pythagoras' theorem its length is √8² + 6² = √100 = 10 m. Check: P(4, 3) is on the path, since 3 × 4 + 4 × 3 = 24, and x = 4 lies between the two ends.

answer(a) 3x + 4y = 24; (b) at (8, 0) on the south fence and (0, 6) on the west fence, and 10 m long

techniqueThe Tangent to a Circle at a Point · Gradients of Parallel and Perpendicular Lines · Point-Slope Form of a Line

examsO-Level · GCSE Higher

In x²a² − y²b² = 1 the minus sign opens the curve into two branches, here the two sides of the wall, and the gap between them is narrowest at y = 0, where x = −a or x = a. Far from the waist the 1 hardly matters, so replacing it by 0 gives the two straight lines, the asymptotes, that the wall approaches. The width at any other height comes from substituting that value of y and solving for x.

1. Compare the equation with x²a² − y²b² = 1. Here a² = 900 and b² = 3600, so a = 30 and b = 60.
2. At the waist y = 0, so x²900 = 1, x² = 900 and x = −30 or x = 30. The wall is 30 m from the axis on each side, so the diameter at the waist is 2 × 30 = 60 m.
3. For the asymptotes, replace the 1 by 0: x²900 = y²3600. Multiply both sides by 3600: y² = 4x², so y = 2x or y = −2x. (a) The waist is 60 m across, and the asymptotes are y = 2x and y = −2x. Far from the waist the wall runs close to these lines, moving about 1 m farther from the axis for every 2 m farther from the waist.
4. At ground level put y = −80: x²900 = 1 + 64003600 = 1 + 169 = 259. Multiply both sides by 900: x² = 2500, so x = 50, taking the positive root for a distance. The diameter at ground level is 2 × 50 = 100 m.
5. At the top put y = 25: x²900 = 1 + 6253600 = 1 + 25144 = 169144. Multiply both sides by 900: x² = 1056.25, so x = 32.5, and the diameter at the top is 2 × 32.5 = 65 m.
6. (b) The tower is 100 m across at ground level and 65 m across at the top. Check at the ground: 2500900 − 64003600 = 259 − 169 = 1.

answer(a) 60 m, and the asymptotes y = 2x and y = −2x; (b) 100 m at ground level and 65 m at the top

techniqueThe Equation of a Hyperbola

examsH2