Parabolas Using Factorisation
Theory
The graph of \(y = ax^2 + bx + c\) is a parabola. When the quadratic factorises, the \(x\)-intercepts can be read straight from the factors. This Year 10 maths page shows how to sketch a parabola from its factorised form: read the concavity from \(a\), find the intercepts, then the axis of symmetry halfway between them and the turning point. It also covers finding the equation from the intercepts and counting \(x\)-intercepts.
A parabola is the graph of a quadratic, \(y = ax^2 + bx + c\) with \(a \ne 0\). It is symmetric about a vertical line called the axis of symmetry.
The vertex (or turning point) is where the axis meets the curve. It is the lowest point (a minimum) when the parabola is concave up, and the highest point (a maximum) when it is concave down.
The sign of \(a\) alone decides the concavity: \(a > 0\) concave up, \(a < 0\) concave down. The constant \(c\) is the \(y\)-intercept, because \(x = 0\) leaves \(y = c\).
The \(x\)-intercepts are the solutions of \(ax^2 + bx + c = 0\). A parabola can have \(2\), \(1\) or \(0\) of them. When the equation factorises as \(y = a(x - p)(x - q)\), the \(x\)-intercepts are \(p\) and \(q\).
In turning-point form, \(y = a(x - h)^2 + k\), the vertex is \((h, k)\): the graph of \(y = ax^2\) moved \(h\) units across and \(k\) units up.
The axis of symmetry of \(y = ax^2 + bx + c\):
If the \(x\)-intercepts are \(p\) and \(q\), the axis is halfway between them:
Turning-point form, vertex \((h, k)\):
How to sketch a parabola by factorising
- Concavity: \(a > 0\) concave up, \(a < 0\) concave down.
- \(y\)-intercept: put \(x = 0\); it is \(c\).
- \(x\)-intercepts: put \(y = 0\). Take out any common factor (including \(-1\)), then factorise and solve.
- Axis: \(x = -\dfrac{b}{2a}\), or the midpoint of the \(x\)-intercepts.
- Vertex: substitute the axis value into the equation to get \(y\).
- Draw a smooth, symmetric curve through the points. Label every intercept and the vertex. Use symmetry for an extra point: the mirror image of the \(y\)-intercept.
Working backwards and with unknowns
- From the intercepts: write \(y = a(x - p)(x - q)\), substitute one more point to find \(a\), then expand.
- Finding \(b\) and \(c\): when \(a = 1\) and the \(x\)-intercepts are known, expand \((x - p)(x - q)\) and compare coefficients.
- Turning point and intercepts given: write \(y = a(x - p)(x - q)\) and substitute the turning point to find \(a\).
- A letter in a factor: for \(y = (x - p)(x - k)\), the axis is \(x = \dfrac{p + k}{2}\); a given axis then fixes \(k\).
- Unknown constant: find the vertex in terms of the constant, then use 'below, on or above the \(x\)-axis' to count the intercepts (Example 4).
- In context: a ball kicked from the origin that lands \(d\) m away has path \(y = ax(x - d)\); its greatest height is at \(x = \dfrac{d}{2}\), which fixes \(a\).
Find the \(y\)-intercept, factorise for the \(x\)-intercepts, then the turning point sits halfway between them.
| \(x = 0:\ y\) | \(=\) | \(-12\) |
| \(x^2 - 4x - 12\) | \(=\) | \(0\) |
| \((x - 6)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(6 \text{ or } x = -2\) |
| \(\text{axis: } x\) | \(=\) | \(\dfrac{-2 + 6}{2} = 2\) |
| \(y\) | \(=\) | \(2^2 - 4(2) - 12\) |
| \(\) | \(=\) | \(\mathbf{-16}\) |
\(a = 1 \gt 0\), so the parabola is concave up and \((2, -16)\) is a minimum. By symmetry, \((4, -12)\) is also on the graph.
Take out the common factor \(-2\) first, so the bracket is monic and easy to factorise.
| \(x = 0:\ y\) | \(=\) | \(10\) |
| \(-2x^2 - 8x + 10\) | \(=\) | \(0\) |
| \(-2(x^2 + 4x - 5)\) | \(=\) | \(0\) |
| \(-2(x + 5)(x - 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-5 \text{ or } x = 1\) |
Use the formula for the axis as a check on the midpoint.
| \(x\) | \(=\) | \(-\dfrac{b}{2a} = -\dfrac{-8}{2 \times (-2)} = -2\) |
| \(y\) | \(=\) | \(-2(-2)^2 - 8(-2) + 10\) |
| \(\) | \(=\) | \(-8 + 16 + 10 = \mathbf{18}\) |
\(a = -2 < 0\): concave down, so \((-2, 18)\) is a maximum turning point.
An \(x\)-intercept at \(-3\) gives the factor \((x + 3)\); at \(1\) it gives \((x - 1)\). The value of \(a\) is still unknown.
| \(y\) | \(=\) | \(a(x + 3)(x - 1)\) |
| \(8\) | \(=\) | \(a(-1 + 3)(-1 - 1)\) |
| \(8\) | \(=\) | \(-4a\) |
| \(a\) | \(=\) | \(-2\) |
Expand:
| \(y\) | \(=\) | \(-2(x^2 + 2x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{-2x^2 - 4x + 6}\) |
Check the \(y\)-intercept: \(-2(3)(-1) = 6\) \(\checkmark\)
(ii) Explain why \(y = x^2 - 2x + 5\) has no \(x\)-intercepts.
(i) The axis is \(x = -\dfrac{8}{2} = -4\). Substitute:
| \(y\) | \(=\) | \((-4)^2 + 8(-4) + c = c - 16\) |
The turning point is \((-4,\ c - 16)\) and the parabola is concave up.
| \(\text{two: below the axis}\) | \(\) | \(c - 16 \lt 0,\ \boldsymbol{c \lt 16}\) |
| \(\text{one: on the axis}\) | \(\) | \(\boldsymbol{c = 16}\) |
| \(\text{none: above the axis}\) | \(\) | \(\boldsymbol{c \gt 16}\) |
(ii) The axis is \(x = 1\) and \(y = 1 - 2 + 5 = 4\), so the turning point is \((1, 4)\). The parabola is concave up, so \(y \ge 4\) for every \(x\): it never reaches the \(x\)-axis. (In completed-square form, \(y = (x - 1)^2 + 4\).)
Common pitfalls
Frequently asked questions
How do you find the vertex of a parabola?
Find the axis of symmetry with \(x = -\dfrac{b}{2a}\), or take the midpoint of the \(x\)-intercepts. Then substitute that \(x\)-value into the equation to find the \(y\)-value of the vertex.
How do you know if a parabola is concave up or down?
Look at \(a\), the coefficient of \(x^2\). If \(a > 0\) the parabola is concave up and has a minimum. If \(a < 0\) it is concave down and has a maximum. The values of \(b\) and \(c\) do not change the concavity.
How do you find the x-intercepts of a parabola?
Put \(y = 0\) and solve the quadratic equation, usually by factorising. Each solution is an \(x\)-intercept. If the equation has no solution, the parabola does not cross the \(x\)-axis.
Can a parabola have no x-intercepts?
Yes. A concave-up parabola whose vertex is above the \(x\)-axis, or a concave-down parabola whose vertex is below it, never meets the \(x\)-axis. For example \(y = x^2 + 4\) has no \(x\)-intercepts.
How do you find the equation of a parabola from its graph?
Read the \(x\)-intercepts \(p\) and \(q\) and write \(y = a(x - p)(x - q)\). Substitute another point on the graph, such as the \(y\)-intercept, to find \(a\). Then expand if the form \(ax^2 + bx + c\) is needed.
What does y = (x - 4) squared + 1 mean?
It is the parabola \(y = x^2\) moved \(4\) units right and \(1\) unit up. Its vertex is \((4, 1)\) and its axis of symmetry is \(x = 4\).