Graphs: linear and quadratic – Year 9 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(x=-2\): \(2(-2)+1=-3\); \(x=-1\): \(-1\); \(x=0\): \(1\); \(x=1\): \(3\); \(x=2\): \(5\); \(x=3\): \(7\). Row: \(-3,-1,1,3,5,7\). Substitute each value of \(x\) into \(2x+1\), using brackets for the negative numbers.
   2. \((0,1)\), where \(x=0\). The line crosses the \(y\)-axis where \(x=0\). The table shows \(y=1\) there, so the point is \((0,1)\).
   3. The gradient is 2: \(y\) goes up by 2 each time \(x\) goes up by 1. Each \(y\)-value is 2 more than the one before, because the coefficient of \(x\) is 2.
   4. \(2\times10+1=\) 21 Substitute \(x=10\) into the equation: \(y=21\).

2. Answer to Question 2

   1. \(m=\) 3, \(c=\) \(-4\) The equation is already in the form \(y=mx+c\): the number in front of \(x\) is the gradient and the number on its own is the intercept.
   2. \(y=-1x+5\), so \(m=\) \(-1\), \(c=\) 5 The term \(-x\) means \(-1\times x\), so the gradient is negative and the line slopes downwards.
   3. \(m=\) \(\dfrac12\), \(c=\) 2 A fractional gradient means the line rises 1 unit for every 2 units across.
   4. Divide by 2: \(y=2x+3\), so \(m=\) 2, \(c=\) 3 The equation must start \(y=\) before you can read \(m\) and \(c\), so divide every term by 2 first.

3. Answer to Question 3

   1. \(m=(7-3)\div2=2\); \(c=3\): \(y=2x+3\) The point \((0,3)\) is on the \(y\)-axis, so the intercept is 3. Check with \((2,7)\): \(2\times2+3=7\).
   2. \(m=(0-5)\div5=-1\); \(c=5\): \(y=-x+5\) The line goes down 5 for every 5 across, so the gradient is \(-1\). Check: \(-5+5=0\).
   3. \(m=(7-3)\div(3-1)=2\). Using \((1,3)\): \(3=2+c\), so \(c=1\): \(y=2x+1\) Neither point is on the \(y\)-axis, so substitute one point into \(y=2x+c\) to find \(c\). Check with \((3,7)\): \(6+1=7\).
   4. \(m=(7+1)\div4=2\); \(c=-1\): \(y=2x-1\) The change in \(y\) is \(7-(-1)=8\) and the change in \(x\) is 4. Check with \((4,7)\): \(8-1=7\).

4. Answer to Question 4

   1. \(3(4)-2=10\). Yes. Substitute \(x=4\): the result equals the \(y\)-coordinate 10, so the point is on the line.
   2. \(3(-1)-2=-5\). Yes. Substitute \(x=-1\): the result equals the \(y\)-coordinate \(-5\), so the point is on the line.
   3. \(3(2)-2=4\), not 5. No. The value of \(y\) given by the equation is 4, but the point has \(y=5\), so it lies just above the line.
   4. Parallel lines have the same gradient 3. \(y=3x+7\) has \(m=3\) (parallel). \(2y=6x+1\) gives \(y=3x+0.5\) (parallel). The others have gradients \(-3\) and \(2\). Always rearrange into \(y=mx+c\) before comparing gradients. Two lines are parallel: \(y=3x+7\) and \(2y=6x+1\).

5. Answer to Question 5

   1. \((-3)^2-4=5\); \(0\); \(-3\); \(-4\); \(-3\); \(0\); \(5\). Row: \(5,0,-3,-4,-3,0,5\). Square \(x\) first and then subtract 4; a negative number squared is positive, so \((-3)^2=9\).
   2. \((0,-4)\) The smallest value in the table is \(-4\), when \(x=0\), so the turning point is a minimum at \((0,-4)\).
   3. \(y=0\) at \(x=-2\) and \(x=2\). The curve crosses the \(x\)-axis where \(y=0\); the table shows this at two values of \(x\).
   4. \(x=0\) (the \(y\)-axis). The table is symmetrical about \(x=0\): for example \(x=-2\) and \(x=2\) both give \(y=0\).

6. Answer to Question 6

   1. \((-2)^2+2(-2)-3=4-4-3=-3\). Matches. Substituting into the equation gives the same value as the table, so the table entry is correct.
   2. \(y=0\) at \(x=-3\) and \(x=1\). The roots are the solutions of \(x^2+2x-3=0\), which are the \(x\)-values where the curve meets the \(x\)-axis.
   3. The minimum is \(-4\), at \(x=-1\). The values decrease and then increase, and the smallest one is \(-4\) at \(x=-1\), so this is the turning point.
   4. \(y=5\) at \(x=-4\) and \(x=2\). Look for the value 5 in the \(y\)-row: it appears at two values of \(x\), one on each side of the line of symmetry.

7. Answer to Question 7

   1. \(3,7,11,15,19\) (since \(2x+3\) for \(x=0,2,4,6,8\)). Substitute each number of miles into \(2x+3\). Each extra 2 miles adds £4 to the fare.
   2. The gradient is the cost per mile (£2) and the intercept is the fixed fee (£3). The intercept is the fare when \(x=0\), before any miles are travelled; the gradient is the extra cost for each mile.
   3. \(2\times7+3=\) £17 Substitute \(x=7\) into the equation. A 7-mile journey costs £17.
   4. \(2x+3=25\), so \(2x=22\) and \(x=\) 11 miles. Set the fare equal to 25 and solve the equation. Sunita can travel 11 miles.
   5. \(x<2x+3\) is always true, so the £1-per-mile firm is cheaper at every distance. The second firm's line \(y=x\) starts lower and rises more slowly, so it is always below \(y=2x+3\) for positive distances.

8. Answer to Question 8

   1. \(15\times6+20=\) £110 Substitute \(x=6\) into the equation. Six months cost £110.
   2. \(15x+20=200\), so \(15x=180\) and \(x=\) 12 months. Set the cost equal to 200, subtract 20 and divide by 15. It takes 12 months.
   3. It is a straight line with gradient 15 that crosses the \(y\)-axis at \((0,20)\): the joining fee. The \(y\)-intercept is the cost when \(x=0\), so it is the joining fee of £20.
   4. \(y=25x\). Equal when \(25x=15x+20\), so \(10x=20\) and \(x=\) 2 months (£50 each). The second gym's line goes through the origin because there is no joining fee. After 2 months both gyms cost £50; after that the first gym is cheaper.

9. Answer to Question 9

   1. \(20-2.5\times3=20-7.5=\) 12.5 cm Substitute \(x=3\) into the formula. After 3 hours the candle is 12.5 cm tall.
   2. The candle burns down by 2.5 cm every hour (negative because the height decreases). The gradient is the change in height per hour; it is negative because the candle gets shorter.
   3. \(20-2.5x=0\), so \(x=20\div2.5=\) 8 hours. The candle has burnt out when its height is 0, so set the formula equal to 0 and solve. It lasts 8 hours.
   4. The end points are \((0,20)\) and \((8,0)\). The line starts at the original height of 20 cm and finishes on the \(x\)-axis when the candle has burnt out.

10. Answer to Question 10

    1. A: \(5+6=\) 11 cm. B: \(8+6=\) 14 cm. Substitute \(x=4\) into both equations. After 4 weeks plant A is 11 cm tall and plant B is 14 cm tall.
    2. Yes, both have gradient 1.5, so both grow 1.5 cm a week and B stays 3 cm taller. Lines with the same gradient are parallel, so the difference in height of 3 cm never changes.
    3. \(5+1.5x=29\), so \(1.5x=24\) and \(x=\) 16 weeks. Set plant A's height equal to 29 and solve the equation. This happens after 16 weeks.
    4. \(2x+5=1.5x+5\) gives \(0.5x=0\), so \(x=0\): they start at the same height and C grows faster, so they are equal only at the start. The two equations give the same height only when \(x=0\); after that plant C grows 0.5 cm a week faster.

11. Answer to Question 11

    1. \(0,5,8,9,8,5,0\) Substitute each value of \(x\) into \(x(6-x)\). For example, when \(x=2\), \(h=2\times4=8\).
    2. The greatest height is 9 m at \(x=3\) m. The largest value in the table is 9, in the middle of the table at \(x=3\), which is the turning point.
    3. \(h=5\) at \(x=1\) and \(x=5\). The value 5 appears twice in the table, at the same distance on either side of \(x=3\).
    4. It lands where \(h=0\) again, at 6 m. The ball starts at height 0 when \(x=0\) and returns to height 0 when \(x=6\), so it lands 6 m away.

12. Answer to Question 12

    1. \(20^2\div20=20\); \(30^2\div20=45\); \(40^2\div20=80\); \(60^2\div20=180\). Square the speed first, then divide by 20. For example, \(30^2=900\) and \(900\div20=45\).
    2. No. Doubling from 20 to 40 mph multiplies the distance by \(80\div20=4\). Doubling the speed multiplies the braking distance by 4, because the speed is squared in the formula.
    3. \(v^2=125\times20=2500\), so \(v=\) 50 mph. Multiply both sides by 20 and then take the positive square root. The speed is 50 mph.
    4. No. The relation involves \(v^2\), so the graph is a curve (a parabola) that gets steeper. Equal increases in speed give bigger and bigger increases in distance, so the points do not lie on a straight line.

13. Answer to Question 13

    1. A: \(0.05\times200+10=20\). B: \(0.1\times200=20\). Substitute \(x=200\) into each equation. Both plans cost £20 for 200 minutes.
    2. The lines cross at (200, 20): 200 minutes costs £20 on both plans. The lines cross where the costs are equal; the tables show this happens at 200 minutes.
    3. At 150 min: A \(=7.5+10=£17.50\), B \(=£15\), so Plan B is cheaper. Substitute \(x=150\) into each equation and compare. Plan B is £2.50 cheaper for 150 minutes.
    4. At 300 min A costs £25 and B costs £30, so Plan A is cheaper. Algebra: \(0.05x+10=0.1x\) gives \(10=0.05x\) and \(x=200\). For more than 200 minutes Plan A is cheaper because its cost per minute is lower. The algebra confirms the crossing at 200 minutes.

14. Answer to Question 14

    1. \(\tfrac12(6)+2=5\). Yes. Substitute \(x=6\) into the equation of the fence; the result equals the \(y\)-coordinate, so the tree is on the fence line.
    2. \(\tfrac12(4)+2=4\), not 3. No. The equation gives \(y=4\) at \(x=4\), but the pond has \(y=3\), so the pond is 1 m below the line.
    3. \(0=\tfrac12x+2\), so \(x=-4\): at \((-4,0)\). The \(x\)-axis is where \(y=0\). Subtract 2 and multiply by 2 to solve the equation.
    4. Parallel means the same gradient \(\tfrac12\) and \(c=-1\): \(y=\tfrac12x-1\). Parallel lines have the same gradient, and the point \((0,-1)\) gives the intercept.

15. Answer to Question 15

    1. Line 1: \(2\), \(4\), \(x\)-int \(-2\). Line 2: \(-3\), \(6\), \(x\)-int \(2\). Line 3: \(\tfrac12\), \(-1\), \(x\)-int \(2\). Line 4: \(4\), \(0\), \(x\)-int \(0\). Read \(m\) and \(c\) from each equation; for the \(x\)-intercept set \(y=0\) and solve the equation.
    2. \(y=4x\), since \(c=0\). A line passes through the origin when its \(y\)-intercept is 0, that is when \(c=0\).
    3. Lines 2 and 3 both cross at \(x=2\). Lines 2 and 3 both have an \(x\)-intercept of 2, so they cross each other on the \(x\)-axis at \((2,0)\).
    4. Through \((2,0)\) with \(m=2\): \(0=4+c\), so \(c=-4\) and \(y=2x-4\). Substitute the point \((2,0)\) into \(y=2x+c\) to find \(c\). Check: \(2\times2-4=0\).

16. Answer to Question 16

    1. Roots \(x=-3\) and \(x=1\). Turning point \((-1,-4)\). The roots are where the curve meets the \(x\)-axis and the turning point is the lowest point of the U.
    2. Line of symmetry \(x=-1\). Mean of roots: \((-3+1)\div2=-1\), the same. The line of symmetry passes through the turning point. It is always halfway between the two roots.
    3. \(y=-3\) at \(x=-2\) and \(x=0\). Draw the horizontal line \(y=-3\) and read the two \(x\)-values where it crosses the curve.
    4. \(0.25+1-3=\) \(-1.75\), close to \(-1.8\) on the graph. Substituting \(x=0.5\) gives \(-1.75\), so the graph reading is a good estimate.
    5. The \(x^2\) term makes \(y\) increase faster as \(x\) moves away from the turning point, so the gradient keeps changing. In a linear graph the gradient is the same everywhere; in a quadratic the gradient changes all the time, which bends the graph into a curve.

17. Answer to Question 17

    1. They cross at \((5,15)\): after 5 weeks both have £15. The point where two graphs cross is where both people have the same amount of money at the same time.
    2. \(2x+5=3x\) gives \(x=5\) and \(y=15\). Set the two expressions equal and solve; substitute \(x=5\) back to find \(y=15\).
    3. Hannah \(=2\times3+5=11\); Tom \(=9\). Hannah has more. Substitute \(x=3\) into both equations: Hannah has £11 and Tom has £9, so Hannah has £2 more.
    4. Hannah £21, Tom £24, so Tom has £3 more. Use the table at 8 weeks. After 5 weeks Tom's faster saving puts him ahead.

18. Answer to Question 18

    1. The two rows are equal at \(x=-1\) (both 1) and at \(x=2\) (both 4). The line and the curve meet where the two rows of the table give the same \(y\)-value.
    2. \((-1,1)\) and \((2,4)\). Each point of intersection uses the \(x\)-value from part a) and the common \(y\)-value.
    3. \((-1)^2=1=-1+2\) and \(2^2=4=2+2\). Substituting each \(x\)-value gives the same number on both sides of the equation, so these are the solutions.
    4. The curve and line meet where \(x^2=x+2\). Rearranging gives \(x^2-x-2=0\), so its solutions are the \(x\)-values where they cross. Subtracting \(x+2\) from both sides of \(x^2=x+2\) gives the quadratic equation, so it has the same solutions.

19. Answer to Question 19

    1. \(8,3,0,-1,0,3,8\) Substitute each value of \(x\) into \(x^2-6x+8\). For example, when \(x=3\), \(y=9-18+8=-1\).
    2. The lowest point is \((3,-1)\): 1 m below ground. The smallest value in the table is \(-1\) at \(x=3\), so the bottom of the ramp is 1 m below the ground.
    3. Between \(x=2\) and \(x=4\) the height is below 0. The height is negative only between the two roots \(x=2\) and \(x=4\).
    4. \(x=3\): for example the values at \(x=2\) and \(x=4\) are both 0. Points at the same distance on either side of \(x=3\) have the same height, for example \(x=1\) and \(x=5\) both give 3.
    5. The graph is symmetrical about \(x=3\), so the left half is the reflection of the right half. A parabola is symmetrical about the vertical line through its turning point, so the two halves are reflections of each other.

20. Answer to Question 20

    1. \(m=(11-5)\div(4-1)=\) 2 Use change in \(y\) divided by change in \(x\) between the two points.
    2. \(5=2(1)+c\), so \(c=3\): \(y=2x+3\). Substitute one of the points into \(y=2x+c\) to find \(c\). Check with \((4,11)\): \(8+3=11\).
    3. \(x=-2,-1,0,1,2,3\) gives \(y=-1,1,3,5,7,9\). Substitute each value of \(x\) into \(y=2x+3\); the values go up by 2 each time.
    4. For \(y=x^2\): \(4,1,0,1,4,9\). They agree at \(x=-1\) (both 1) and \(x=3\) (both 9). Points: \((-1,1)\) and \((3,9)\). Compare the two tables: the line and curve meet where the values are equal.
    5. \(x^2=2x+3\) gives \(x^2-2x-3=0\). \((x-3)(x+1)=x^2+x-3x-3=x^2-2x-3\). So \(x=3\) or \(x=-1\). Expanding the brackets gives back the quadratic, so its solutions \(x=3\) and \(x=-1\) agree with the points found from the tables.