Home / iGCSE Mathematics (0580) :2.11Recognise, sketch and interpret graphs of functions. iGCSE Style Questions Paper 3

iGCSE Mathematics (0580) :2.11Recognise, sketch and interpret graphs of functions. iGCSE Style Questions Paper 3

Question

(a)

(i) Complete the table of values for \(y = x^2 + x\).

(ii) On the grid, draw the graph of \( y = x^2 + x\) for \(-3 \leq x \leq 3\).

(iii) On the grid, draw the line y = 10.
(iv) Use both your graphs to solve \(x^2 + x = 10\) for \(-3 \leq x \leq 3\).

(b) Another line, L, has the equation \(y = \frac{2}{3}x -5\).
(i) Write down the gradient of L.
(ii) Write down the equation of a straight line that is parallel to L.

(c)

Write the equation of the line, K, in the form y = mx + c.

▶️ Answer/Explanation
Solution

(a)

(i) Ans: 2 and 2, 12

For \(x = 1\), \(y = 1^2 + 1 = 2\). For \(x = 2\), \(y = 2^2 + 2 = 6\). For \(x = 3\), \(y = 3^2 + 3 = 12\).

(ii)

Plot points \((-3,6), (-2,2), (-1,0), (0,0), (1,2), (2,6), (3,12)\) and draw a smooth curve.

(iii)

Draw a horizontal line at \(y = 10\).

(iv) Ans: 2.6 – 2.8

Find intersection points of \(y = x^2 + x\) and \(y = 10\). Approximate solutions are \(x \approx 2.7\) and \(x \approx -3.7\) (only \(x \approx 2.7\) is valid in \(-3 \leq x \leq 3\)).

(b)

(i) Ans: \(\frac{2}{3}\)

The gradient of \(L\) is the coefficient of \(x\) in \(y = \frac{2}{3}x -5\).

(ii) Ans: \(y = \frac{2}{3}x + c\)

Parallel lines have the same gradient, so any line of the form \(y = \frac{2}{3}x + c\) is parallel to \(L\).

(c) Ans: \(y = 2x – 3\)

From the graph, the line \(K\) passes through \((0,-3)\) and \((1,-1)\). The gradient \(m = \frac{-1 – (-3)}{1 – 0} = 2\). Thus, the equation is \(y = 2x – 3\).

Question

 (a) Complete the table of values for \(y = 8 + 7x − x^2\).

(b) On the grid, draw the graph of \(y = 8 + 7x − x^2\) for \(0 \leq x \leq 8\).

(c) Write down the co-ordinates of the highest point of the curve.

(d) (i) On the grid, draw the line y = 16.
(ii) Use your line to solve the equation \(8 + 7x − x^2 = 16\).

▶️ Answer/Explanation
Solution

(a) Ans: … 14 … 20 20 … 14 … 0

Substitute x-values into \(y = 8 + 7x – x^2\): For x=2, y=8+14-4=18; x=3, y=8+21-9=20; x=6, y=8+42-36=14; x=8, y=8+56-64=0.

(b)

Plot all points from the table and join them with a smooth curve. The parabola should pass through (0,8), (1,14), (2,18), (3,20), etc.

(c) Ans: (3.5, 20.25)

The vertex (highest point) occurs at \(x = -b/2a = -7/(2×-1) = 3.5\). Substitute back to get y=20.25.

(d)(i)

Draw a horizontal line through y=16 across the grid.

(d)(ii) Ans: 1.4, 5.6

Find where the curve intersects y=16. The solutions are approximately x=1.4 and x=5.6.

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