iGCSE Mathematics (0580) - C2.10 Graphs of functions- Exam Style Questions Paper 1- New Syllabus
▶️ Answer/Explanation
(a)
Substitute the two missing \(x\) values into \(y = \dfrac{9}{x}\):
When \(x = -2\): \(y = \dfrac{9}{-2} = -4.5\)
When \(x = 3\): \(y = \dfrac{9}{3} = 3\)
✅ Answer: \(-4.5\) and \(3\)
(b)
Plot all the points from the completed table on the grid.
The graph of \(y = \dfrac{9}{x}\) is a reciprocal curve (hyperbola) in two separate branches — one in the third quadrant (negative \(x\) and negative \(y\)) and one in the first quadrant (positive \(x\) and positive \(y\)).
Join the points with a smooth curve in each branch; do not connect the two branches or cross the axes.
✅ Answer: Smooth correct curve through all plotted points. (B3FT for 9 points correct; B2FT for 7; B1FT for 5.)
▶️ Answer/Explanation
(a) To fill in the missing $y$ values, divide 12 by each corresponding $x$ coordinate:
For $x = -6$, $y = \frac{12}{-6} = -2$; for $x = -3$, $y = \frac{12}{-3} = -4$; for $x = -1$, $y = \frac{12}{-1} = -12$;
For $x = 1$, $y = \frac{12}{1} = 12$; for $x = 3$, $y = \frac{12}{3} = 4$; for $x = 6$, $y = \frac{12}{6} = 2$.
(b) Plot these calculated points on your axes grid and connect them with two smooth, separate curves across opposite quadrants.
(c) Draw a completely straight, flat horizontal line passing through $-9$ on the vertical $y$-axis.
(d) Look at where your curve intersects the horizontal line $y = -9$ and read the value straight up to the $x$-axis, which should land between $-1.4$ and $-1.2$.
✅ Answer:
(a) Missing values are: -2, -4, -12, 12, 4, 2 respectively
(b) Correct curve drawn
(c) Line $y = -9$ drawn correctly
(d) $x$ value in the range -1.4 to -1.2