How to Plot Quadratic Graphs (y = ax² + bx + c) | IGCSE Maths
Quadratic graphs produce smooth, symmetrical curves called parabolas. On this page you will learn how to plot graphs of the form y = ax2 + bx + c using a table of values, recognise whether the parabola opens upwards or downwards, and read solutions directly from the graph. These skills are essential for the IGCSE Maths exam and provide the foundation for later work on turning points and graphical transformations.
Core Ideas
The shape of a parabola
The sign of \(a\) (the number in front of \(x^2\)) controls which way the curve faces. Think of it this way: a positive \(a\) gives a smiley face (U-shaped), and a negative \(a\) gives a sad face (inverted U).
Positive \(a\): U-shaped 😊
The curve opens upwards.
It has a minimum turning point.
Negative \(a\): Inverted U 😞
The curve opens downwards.
It has a maximum turning point.
How to plot a quadratic graph
The method is the same every time: build a table of values, plot the points, and draw a smooth curve through them. There are two ways to fill in the table, and you should be comfortable with both.
💡 Example 1: Plot \(y = x^2 + 3x - 4\) for \(-5 \leq x \leq 2\)
Method A: Direct substitution
Substitute each \(x\)-value directly into the equation. For example, when \(x = -3\):
Repeating for every \(x\)-value gives:
| \(x\) | \(-5\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 6 | 0 | \(-4\) | \(-6\) | \(-6\) | \(-4\) | 0 | 6 |
Method B: Split into separate rows
This method breaks the equation into its individual terms. It is slower but makes sign errors much easier to spot, especially when \(a \neq 1\) or when substituting negative values.
| \(x\) | \(-5\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|---|
| \(x^2\) | 25 | 16 | 9 | 4 | 1 | 0 | 1 | 4 |
| \(3x\) | \(-15\) | \(-12\) | \(-9\) | \(-6\) | \(-3\) | 0 | 3 | 6 |
| \(-4\) | \(-4\) | \(-4\) | \(-4\) | \(-4\) | \(-4\) | \(-4\) | \(-4\) | \(-4\) |
| \(y\) | 6 | 0 | \(-4\) | \(-6\) | \(-6\) | \(-4\) | 0 | 6 |
Plot the points and draw a smooth curve
The curve is U-shaped (smiley face) because \(a = 1 > 0\). It crosses the \(x\)-axis at \(x = -4\) and \(x = 1\), so the solutions of \(x^2 + 3x - 4 = 0\) are \(x = -4\) and \(x = 1\).
💡 Example 2: Plot \(y = -2x^2 + 12x - 10\) for \(0 \leq x \leq 6\)
Direct substitution check: when \(x = 2\)
Full table (split into rows, useful when \(a \neq 1\))
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|---|
| \(-2x^2\) | 0 | \(-2\) | \(-8\) | \(-18\) | \(-32\) | \(-50\) | \(-72\) |
| \(12x\) | 0 | 12 | 24 | 36 | 48 | 60 | 72 |
| \(-10\) | \(-10\) | \(-10\) | \(-10\) | \(-10\) | \(-10\) | \(-10\) | \(-10\) |
| \(y\) | \(-10\) | 0 | 6 | 8 | 6 | 0 | \(-10\) |
Plot and draw the curve
The curve is an inverted U (sad face) because \(a = -2 < 0\). It crosses the \(x\)-axis at \(x = 1\) and \(x = 5\), so the solutions of \(-2x^2 + 12x - 10 = 0\) are \(x = 1\) and \(x = 5\). The turning point (maximum) is at \((3, 8)\).
Using a quadratic graph to solve an equation
Once you have drawn the graph of \(y = ax^2 + bx + c\), the solutions of \(ax^2 + bx + c = 0\) are found where the curve crosses the \(x\)-axis (because that is where \(y = 0\)).
A parabola can cross the \(x\)-axis twice (two solutions), touch it once (one repeated solution), or miss it entirely (no real solutions).
💡 Example 3: Two solutions
The graph of \(y = x^2 - 6x + 8\) is shown below. Use it to solve \(x^2 - 6x + 8 = 0\).
The curve crosses the \(x\)-axis at \(x = 2\) and \(x = 4\).
\[x = 2 \quad \text{or} \quad x = 4\]Check: \(2^2 - 6(2) + 8 = 0\) ✓ and \(4^2 - 6(4) + 8 = 0\) ✓
💡 Example 4: One repeated root
The graph of \(y = x^2 - 4x + 4\) just touches the \(x\)-axis. Solve \(x^2 - 4x + 4 = 0\).
The curve touches the \(x\)-axis at \(x = 2\) only.
\[x = 2 \quad \text{(repeated root)}\]This happens when the vertex sits exactly on the \(x\)-axis. The quadratic factorises as \((x - 2)^2 = 0\).
Real-life quadratic graphs
Quadratic equations appear in many practical situations. The variables \(x\) and \(y\) are replaced with quantities like length, area, time, or height. A common exam question involves finding the dimensions that give the maximum area for a fixed perimeter.
💡 Example 5: Maximum area problem
A rectangular flower bed is to be edged with 24 m of border. If the width of the bed is \(x\) m, find the dimensions that give the largest area.
All four sides are edged, so the two widths and two lengths add up to 24 m: \(2x + 2L = 24\), giving \(L = 12 - x\).
Length \(= 12 - x\)
Area:
\[\begin{array}{rcl} A &=& x(12 - x) \\ &=& 12x - x^2 \end{array}\]Construct a table of values and plot the graph
| \(x\) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| \(12x\) | 0 | 24 | 48 | 72 | 96 | 120 | 144 |
| \(-x^2\) | 0 | \(-4\) | \(-16\) | \(-36\) | \(-64\) | \(-100\) | \(-144\) |
| \(A\) | 0 | 20 | 32 | 36 | 32 | 20 | 0 |
From the graph, the maximum area is 36 m², which occurs when \(x = 6\).
The length is \(12 - 6 = 6\) m, so the dimensions are 6 m by 6 m (a square).
The curve is an inverted U (sad face) because the coefficient of \(x^2\) is negative (\(-1\)). The turning point gives the maximum value. Notice that \(x\) must be between 0 and 12 for the dimensions to make sense, and that for a fixed perimeter the largest rectangle is always a square.
🔑 Key Points
A quadratic equation has the form \(y = ax^2 + bx + c\). The graph is always a parabola.
If \(a > 0\): U-shaped, smiley face 😊 (minimum). If \(a < 0\): inverted U, sad face 😞 (maximum).
The roots (solutions of \(ax^2 + bx + c = 0\)) are the \(x\)-values where the curve crosses the \(x\)-axis.
Always break the table into separate rows for each term (\(ax^2\), \(bx\), \(c\)) to reduce arithmetic errors.
Join the plotted points with a smooth curve, never with straight-line segments.
Real-life problems may use different letters (e.g. \(A\) and \(x\)) instead of \(y\) and \(x\).
⚠️ Common Mistakes
Sign errors when squaring negatives: \((-3)^2 = 9\), not \(-9\).
Forgetting \(a\): for \(2x^2\), you need \(2 \times (-3)^2 = 18\), not \((-3)^2 = 9\).
Joining with straight lines: a parabola is always a smooth curve. Straight segments between points will lose you marks.
Too few points: you need enough points (at least 6 or 7) to see where the curve turns. If in doubt, calculate extra values near the turning point.
Misreading roots: the roots are where the curve crosses the \(x\)-axis, not the \(y\)-axis.