IGCSE Mathematics: Graphs of Functions Practice Questions
A quadratic graph is a parabola whose roots are found by factorising and whose y-intercept is the constant term. Solving an equation graphically means drawing a second line and reading off the x values where the graphs cross.
Topic 2.10 of Cambridge IGCSE Mathematics 0580 combines sketching with reading. The graphical solution method is worth learning precisely, because Cambridge specifies which line must be drawn and awards a mark for it. The questions below cover both skills.
What you need to know for Graphs of Functions
- Quadratic graphA parabola. It opens upwards when the coefficient of x2 is positive and downwards when it is negative.
- Roots and interceptsThe roots are where y = 0, found by factorising and solving. The y-intercept is the value of y when x = 0, which is the constant term.
- Line of symmetryA parabola is symmetrical about a vertical line halfway between its two roots. The turning point lies on that line.
- Completed square formy = (x minus p)2 + q has its turning point at (p, q), a minimum when the squared term is positive.
- Reciprocal graphy = 1 divided by x has two separate branches and never touches either axis. The axes are asymptotes.
- Solving graphicallyRearrange the equation so one side matches the curve already drawn. Draw the other side as a new line and read the x values at the intersections.
IGCSE Mathematics Graphs of Functions questions and answers
4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
For the curve y = x2 minus 4x + 3, find the coordinates of the two roots, the y-intercept and the turning point.
Show the worked answer
- For the roots, set y = 0 and factorise: x2 minus 4x + 3 = (x minus 1)(x minus 3) = 0, so x = 1 or x = 3.
- The roots are therefore at (1, 0) and (3, 0).
- For the y-intercept, set x = 0: y = 3, giving the point (0, 3).
- The line of symmetry lies halfway between the roots, at x = (1 + 3) divided by 2 = 2.
- Substitute x = 2 to find the y value of the turning point: y = 4 minus 8 + 3 = minus 1, so the turning point is (2, minus 1). It is a minimum, since the coefficient of x2 is positive.
The graph of y = x2 minus 4x + 3 has been drawn. Describe how to use it to solve x2 minus 4x + 1 = 0.
Show the worked answer
- The drawn curve is y = x2 minus 4x + 3, so the equation must be rearranged to make one side match that expression.
- Starting from x2 minus 4x + 1 = 0, add 2 to both sides: x2 minus 4x + 3 = 2.
- The left hand side is now exactly the curve, so the solutions occur where the curve equals 2.
- Draw the horizontal line y = 2 on the same axes and read off the x coordinates of the two points where it crosses the curve. Those x values are the solutions.
Describe the shape of the graph of y = 1 divided by x, and explain why the curve never touches either axis.
Show the worked answer
- The graph has two separate branches. One lies in the first quadrant where both x and y are positive, and the other in the third quadrant where both are negative.
- As x becomes very large, 1 divided by x becomes very small but never reaches zero, so the curve approaches the x-axis without touching it.
- As x approaches zero, 1 divided by x becomes very large, so the curve approaches the y-axis without touching it.
- x = 0 is not in the domain at all, because division by zero is undefined. Both axes are therefore asymptotes.
Write down the coordinates of the turning point of y = (x minus 3)2 + 4 and state whether it is a maximum or a minimum.
Show the worked answer
- The expression is in completed square form, y = (x minus p)2 + q, whose turning point is at (p, q).
- Here p = 3, taken from inside the bracket with the sign reversed, and q = 4.
- The turning point is therefore at (3, 4).
- A squared term is never negative, so the smallest possible value of (x minus 3)2 is zero, which happens at x = 3. The smallest value of y is therefore 4, making the turning point a minimum.
Common mistakes in this topic
- Reading the sign of the turning point from a completed square without reversing it.
- Drawing the wrong line when solving graphically.
- Joining the two branches of a reciprocal graph.
- Giving roots as values when coordinates are asked for.
- Assuming a parabola opens upwards without checking the sign of the x squared coefficient.
Exam tips
- Find the roots, the y-intercept and the line of symmetry before sketching. Those three pieces fix the whole curve.
- The line of symmetry always sits halfway between the two roots.
- For graphical solutions, rearrange until the left hand side matches the printed curve exactly, then draw whatever is left on the right.
- State the equation of the line you draw. Cambridge awards a mark for it specifically.
- Use a sharp pencil and draw a smooth curve, never a series of straight segments between plotted points.
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Graphs of Functions FAQs
How do I find the turning point of a quadratic?
Find the two roots by factorising and take the value halfway between them, which gives the x coordinate of the turning point. Substitute that back into the equation for the y coordinate. Alternatively, if the equation is in completed square form, the turning point can be read straight off.
How do I solve an equation using a graph that is already drawn?
Rearrange your equation so that one side is exactly the expression that was plotted. Whatever remains on the other side is the line you must draw. Read the x coordinates where that line crosses the curve, and those are your solutions.
Why does the graph of y equals 1 over x never touch the axes?
As x becomes very large, y becomes very small but never actually reaches zero, so the curve approaches the x-axis without meeting it. As x approaches zero, y becomes very large, and at x equals zero the function is undefined because division by zero is not allowed.
How do I read the turning point from completed square form?
For y equals (x minus p) squared plus q, the turning point is at the point (p, q). The sign inside the bracket reverses, so (x minus 3) squared plus 4 has its turning point at (3, 4). A positive squared term means it is a minimum.
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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.