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

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1. Answer to Question 1 (4 marks)

   1. Substitute each value of x into \(3x-2\): \(x=-1\) gives \(-3-2=-5\); \(x=0\) gives \(-2\); \(x=1\) gives 1; \(x=2\) gives 4; \(x=3\) gives 7. The values are \(-5,-2,1,4,7\) (2 marks for all five, 1 mark for at least three).
   2. In \(y=mx+c\) the gradient is the coefficient of x and the intercept is the constant term. Gradient 3 (1 mark), y-intercept \(-2\) (1 mark).

   The table confirms the gradient: each y value is 3 more than the one before as x increases by 1. It also confirms the intercept: when \(x=0\), \(y=-2\), so the line crosses the y-axis at \((0,-2)\). Plot the points and join them with a ruler.

2. Answer to Question 2 (4 marks)

   1. Substitute carefully, using brackets for negatives. \(x=-2\): \((-2)^{2}-2(-2)-3=4+4-3=\)5. \(x=1\): \(1-2-3=\)\(-4\). \(x=4\): \(16-8-3=\)5. (2 marks for all three, 1 mark for two)
   2. The roots are the x-values where the curve crosses the x-axis, so where \(y=0\) in the table: \(x=-1\) and \(x=3\). (1 mark)
   3. A parabola is symmetrical, so the turning point is halfway between the roots, at \(x=\dfrac{-1+3}{2}=1\), where \(y=-4\). The turning point is \((1,-4)\), a minimum. (1 mark)

   Check (b) by factorising: \(x^{2}-2x-3=(x+1)(x-3)\).

3. Answer to Question 3 (4 marks)

   1. Gradient = change in y ÷ change in x. The y-values change by \(4-(-2)=6\) and the x-values by \(3-0=3\), so the gradient is \(6\div3=\)2. (1 mark)
   2. The line passes through \((0,-2)\), so the y-intercept is \(-2\). The equation is \(y=2x-2\). (1 mark)
   3. Substitute \(x=5\): \(2\times5-2=10-2=8\). This gives \(y=8\), matching the point, so \((5,8)\) lies on the line. (1 mark)
   4. Parallel lines have equal gradients, so keep the gradient 2 and use the new intercept 5: \(y=2x+5\). (1 mark)

   Check (b) with the second point: \(2\times3-2=4\). The two parallel lines are always 7 units apart vertically and never meet.

4. Answer to Question 4 (4 marks)

   1. In each equation the monthly charge is the gradient and the joining fee is the intercept. Gym A: \(y=12x+20\) (1 mark). Gym B: \(y=10x+30\) (1 mark).
   2. The costs are equal when \(12x+20=10x+30\) (1 mark for setting them equal). Subtract \(10x\): \(2x+20=30\). Subtract 20: \(2x=10\), so \(x=\)5 months. The cost is \(12\times5+20=\)£80 (1 mark).

   Check with Gym B: \(10\times5+30=80\). On a graph the two lines cross at \((5,80)\). Before 5 months Gym A is cheaper because of its lower joining fee; after 5 months Gym B is cheaper because each month costs £2 less. For a one-year membership Gym B saves £14.

5. Answer to Question 5 (4 marks)

   1. Substitute into \(h=x(8-x)\). For \(x=2\): \(2\times(8-2)=2\times6=\)12 (1 mark). For \(x=6\): \(6\times(8-6)=6\times2=\)12 (1 mark).
   2. The table is symmetrical about \(x=4\), and the largest value sits in the middle. The maximum height is 16 m at a distance of 4 m. (1 mark)
   3. The ball lands when its height returns to 0, which happens at \(x=8\). It travels 8 m before landing. (1 mark)

   Expanding gives \(h=8x-x^{2}\); the negative \(x^{2}\) term makes the curve an upside-down parabola, which has a maximum rather than a minimum. The roots \(x=0\) and \(x=8\) are the throw and the landing, and the maximum is halfway between them.

Total : 20 marks