Sketching Parabolas Using Intercept Form
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Theory
A parabola written as \(y = a(x - p)(x - q)\) is in intercept form: its \(x\)-intercepts are \(p\) and \(q\). This Year 9 maths page shows how to read the intercepts, find the axis of symmetry halfway between them, find the vertex by substituting, and sketch the parabola. It also shows how to factorise first and how to find the equation of a parabola from its intercepts.
The intercept form (or factorised form) of a quadratic is \(y = a(x - p)(x - q)\), with \(a \ne 0\).
On the \(x\)-axis \(y = 0\), so one of the brackets must be zero (the null factor law). The \(x\)-intercepts are therefore \(x = p\) and \(x = q\). Watch the signs: \(y = (x + 3)(x - 5)\) crosses the \(x\)-axis at \(x = -3\) and \(x = 5\).
A parabola is symmetric, so the axis of symmetry is the vertical line halfway between the \(x\)-intercepts. The vertex (turning point) lies on this line: substitute the axis value into the equation to find its \(y\)-value.
The \(y\)-intercept comes from putting \(x = 0\): \(y = a(0 - p)(0 - q) = apq\).
The sign of \(a\) gives the shape. If \(a > 0\) the parabola is concave up and the vertex is a minimum. If \(a < 0\) it is concave down and the vertex is a maximum.
Intercept form, and the features you can read from it:
From the expanded form \(y = ax^2 + bx + c\), the axis is also given by
How to sketch a parabola in intercept form
- Factorise first if the equation is expanded. Take out any common factor (including \(-1\) or another negative number), then factorise the bracket.
- Shape: \(a > 0\) concave up, \(a < 0\) concave down.
- \(x\)-intercepts: set each bracket equal to zero.
- Axis of symmetry: \(x =\) the midpoint of the \(x\)-intercepts.
- Vertex: substitute the axis value into the equation to get the \(y\)-value.
- \(y\)-intercept: put \(x = 0\) (it equals \(apq\)).
- Draw a smooth curve, symmetric about the axis, through the labelled points. The mirror image of the \(y\)-intercept gives one more point.
Working backwards
- From the intercepts and a point: write \(y = a(x - p)(x - q)\), substitute the point to find \(a\), then expand if needed.
- If the equation is not in intercept form (for example \(y = (x - 3)^2 - 1\)), put \(y = 0\) and solve. If there is no solution, there are no \(x\)-intercepts.
- Areas: a triangle with two vertices on the \(x\)-intercepts and one at the vertex has base = the distance between the intercepts and height = the size of the vertex's \(y\)-value.
Each bracket is zero at an \(x\)-intercept. The axis is halfway between them.
| \((x + 3)(x - 5)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-3 \text{ or } x = 5\) |
| \(\text{axis: } x\) | \(=\) | \(\dfrac{-3 + 5}{2} = 1\) |
| \(y\) | \(=\) | \((1 + 3)(1 - 5)\) |
| \(\) | \(=\) | \(\mathbf{-16}\) |
| \(x = 0:\ y\) | \(=\) | \((3)(-5) = -15\) |
\(a = 1 > 0\), so the parabola is concave up and the vertex \((1, -16)\) is a minimum.
Take out the common factor \(-2\) so the bracket is easy to factorise.
| \(y\) | \(=\) | \(-2(x^2 + 2x - 8)\) |
| \(\) | \(=\) | \(-2(x + 4)(x - 2)\) |
| \(x\text{-intercepts: } x\) | \(=\) | \(-4 \text{ or } x = 2\) |
| \(\text{axis: } x\) | \(=\) | \(\dfrac{-4 + 2}{2} = -1\) |
| \(y\) | \(=\) | \(-2(-1 + 4)(-1 - 2)\) |
| \(\) | \(=\) | \(-2(3)(-3) = \mathbf{18}\) |
| \(x = 0:\ y\) | \(=\) | \(16\) |
Check the axis with \(x = -\dfrac{b}{2a} = -\dfrac{-4}{2 \times (-2)} = -1\). Since \(a = -2 < 0\), the parabola is concave down and \((-1, 18)\) is a maximum.
(ii) The parabola \(y = x^2 + bx + c\) has \(x\)-intercepts \(4\) and \(-1\). Find \(b\) and \(c\).
(i) The intercepts give the factors \((x + 3)\) and \((x - 1)\). The value of \(a\) is unknown, so substitute the point.
| \(y\) | \(=\) | \(a(x + 3)(x - 1)\) |
| \(-12\) | \(=\) | \(a(-1 + 3)(-1 - 1)\) |
| \(-12\) | \(=\) | \(-4a\) |
| \(a\) | \(=\) | \(3\) |
| \(y\) | \(=\) | \(3(x^2 + 2x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{3x^2 + 6x - 9}\) |
Check the \(y\)-intercept: \(apq = 3 \times (-3) \times 1 = -9\) \(\checkmark\)
(ii) Here \(a = 1\), so expand \((x - 4)(x + 1)\) and compare.
| \(y\) | \(=\) | \((x - 4)(x + 1)\) |
| \(\) | \(=\) | \(x^2 - 3x - 4\) |
So \(\boldsymbol{b = -3}\) and \(\boldsymbol{c = -4}\).
(i) Find the coordinates of \(A\), \(B\) and \(V\).
(ii) Find the area of triangle \(ABV\).
(iii) The parabola passes through \((0, 24)\). Use symmetry to find another point with \(y = 24\).
(i) Take out \(-1\) and factorise.
| \(y\) | \(=\) | \(-(x^2 + 2x - 24)\) |
| \(\) | \(=\) | \(-(x + 6)(x - 4)\) |
| \(x\) | \(=\) | \(-6 \text{ or } x = 4\) |
| \(\text{axis: } x\) | \(=\) | \(\dfrac{-6 + 4}{2} = -1\) |
| \(y\) | \(=\) | \(-(-1 + 6)(-1 - 4) = 25\) |
So \(\mathbf{A(-6, 0)}\), \(\mathbf{B(4, 0)}\) and \(\mathbf{V(-1, 25)}\).
(ii) The base \(AB\) is \(4 - (-6) = 10\) units and the height is the vertex's \(y\)-value, \(25\).
| \(\text{Area}\) | \(=\) | \(\tfrac{1}{2} \times 10 \times 25\) |
| \(\) | \(=\) | \(\mathbf{125} \textbf{ square units}\) |
(iii) \((0, 24)\) is \(1\) unit right of the axis \(x = -1\), so its mirror image is \(1\) unit left: \(\mathbf{(-2, 24)}\). Check: \(-(-2)^2 - 2(-2) + 24 = -4 + 4 + 24 = 24\) \(\checkmark\)
Common pitfalls
Frequently asked questions
What is the intercept form of a parabola?
It is \(y = a(x - p)(x - q)\). It shows the \(x\)-intercepts straight away: the parabola crosses the \(x\)-axis at \(x = p\) and \(x = q\).
How do you find the axis of symmetry from the x-intercepts?
Add the two \(x\)-intercepts and divide by \(2\). For intercepts \(-2\) and \(6\), the axis is \(x = 2\).
How do you find the vertex from intercept form?
Find the axis of symmetry, then substitute that \(x\)-value into the equation. For \(y = (x - 1)(x - 5)\) the axis is \(x = 3\) and \(y = (2)(-2) = -4\), so the vertex is \((3, -4)\).
How do you find the equation of a parabola from its x-intercepts?
Write \(y = a(x - p)(x - q)\) using the intercepts \(p\) and \(q\). Then substitute one more point on the graph, such as the \(y\)-intercept, to find \(a\).
What if the quadratic is not factorised?
Take out any common factor and factorise first. If \(y = 0\) has no solution, the parabola has no \(x\)-intercepts. For example \(y = x^2 + 4\) is never less than \(4\).