Exploring Parabolas
A parabola is the graph of a quadratic such as \(y=ax^2+bx+c\). It has a vertex, an axis of symmetry, a \(y\)-intercept and up to two \(x\)-intercepts, and it is concave up when \(a>0\) and concave down when \(a<0\). In NSW Year 10 Advanced maths you find these features algebraically, without graphing software (NESA outcome MA5‑NLI‑P‑01).
You will find intercepts by factorising or with the quadratic formula, find the axis of symmetry from \(x=-\dfrac{b}{2a}\) or from the intercepts, and find the vertex and whether it is a maximum or minimum. You will complete the square to find a minimum value and a range, use the vertex to count \(x\)-intercepts, and find the equation of a parabola from its features. Harder questions prove the formula for the axis of symmetry. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Exploring Parabolas - Video - Visual introduction to parabolas Watch
Theory
The graph of \(y = ax^2 + bx + c\) is a parabola. Its key features are the intercepts, the axis of symmetry \(x = -\dfrac{b}{2a}\), the vertex (turning point) and the concavity. This Year 10 Advanced maths page shows how to find each feature, complete the square, find the equation of a parabola and where a line meets it.
A parabola is the graph of a quadratic \(y = ax^2 + bx + c\) with \(a \ne 0\). If \(a \gt 0\) it is concave up (opens upwards); if \(a \lt 0\) it is concave down.
The \(x\)-intercepts are where \(y = 0\); a parabola can have two, one or none. The \(y\)-intercept is where \(x = 0\), which is always \(y = c\).
The axis of symmetry is the vertical line that folds the parabola onto itself. The vertex (or turning point) lies on it: a minimum if concave up, a maximum if concave down. The \(y\)-value of the vertex is the least (or greatest) value of \(y\), and it fixes the range.
The form \(y = a(x - h)^2 + k\) is the turning-point form: the vertex is \((h,\ k)\).
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
Axis of symmetry and vertex of \(y = ax^2 + bx + c\):
Turning-point form and the quadratic formula for the \(x\)-intercepts:
Factored form: \(y = a(x - p)(x - q)\) has \(x\)-intercepts \(p\) and \(q\), axis \(x = \dfrac{p + q}{2}\), and \(y\)-intercept \(apq\).
How to sketch a parabola
- Concavity: read the sign of \(a\).
- \(y\)-intercept: \(y = c\).
- \(x\)-intercepts: let \(y = 0\); factorise, or use the quadratic formula (round as asked).
- Axis and vertex: \(x = -\dfrac{b}{2a}\) or the midpoint of the \(x\)-intercepts; substitute to get the \(y\)-value. Completing the square gives the vertex directly.
- Sketch a smooth curve through the labelled points, symmetric about the axis.
Problem types
- Completing the square with \(a \ne 1\): take out \(a\) from the \(x^2\) and \(x\) terms first (Example 2). With \(a = -1\), take out \(-1\): \(-x^2 + 4x = -(x^2 - 4x)\).
- Matching a graph: eliminate using the sign of \(a\), the \(y\)-intercept \(c\), then the side of the \(y\)-axis the vertex is on.
- Finding the equation: from the vertex use \(y = a(x - h)^2 + k\); from the \(x\)-intercepts use \(y = a(x - p)(x - q)\); then substitute one more point to find \(a\). From three points, substitute each into \(y = ax^2 + bx + c\) and solve the equations simultaneously.
- Line and parabola: set the expressions equal and solve the quadratic. Two solutions: the line cuts twice. One (a repeated root): the line is a tangent. None: they do not meet.
- Unknown coefficient: complete the square with the letter in place, e.g. \(x^2 + kx + 9 = \left(x + \dfrac{k}{2}\right)^2 + 9 - \dfrac{k^2}{4}\), then apply the condition (vertex on the axis, vertex on a given line, and so on).
- Showing there are no \(x\)-intercepts: complete the square to get \((\ldots)^2 + \text{positive}\), which can never be \(0\).
\(x\)-intercepts: let \(y = 0\) and factorise.
| \(x^2 - 2x - 8\) | \(=\) | \(0\) |
| \((x - 4)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(4\ \text{or}\ -2\) |
\(y\)-intercept: let \(x = 0\), so \(y = -8\). Axis: halfway between the intercepts, \(x = \dfrac{-2 + 4}{2} = 1\) (or \(x = -\dfrac{b}{2a} = \dfrac{2}{2} = 1\)).
| \(y\) | \(=\) | \(1^2 - 2(1) - 8 = -9\) |
Vertex \((1,\ -9)\). Since \(a = 1 \gt 0\) the parabola is concave up, so the vertex is a minimum turning point.
Take out the coefficient of \(x^2\) from the first two terms, then add and subtract the square of half the \(x\)-coefficient inside the bracket.
| \(y\) | \(=\) | \(2(x^2 + 6x) + 23\) |
| \(\) | \(=\) | \(2(x^2 + 6x + 9 - 9) + 23\) |
| \(\) | \(=\) | \(2(x + 3)^2 - 18 + 23\) |
| \(\) | \(=\) | \(2(x + 3)^2 + 5\) |
\(2(x + 3)^2 \ge 0\), so \(y \ge 5\), with \(y = 5\) when \(x = -3\).
Vertex \(\mathbf{(-3,\ 5)}\), a minimum. Range \(\boldsymbol{y \ge 5}\). The lowest point is above the \(x\)-axis and the curve is concave up, so there are no \(x\)-intercepts.
Check with \(x = -\dfrac{b}{2a} = -\dfrac{12}{4} = -3\) and \(y = 2(9) - 36 + 23 = 5\).
Start from the turning-point form \(y = a(x - 3)^2 - 8\) and substitute the point \((1,\ 0)\):
| \(0\) | \(=\) | \(a(1 - 3)^2 - 8\) |
| \(4a\) | \(=\) | \(8\) |
| \(a\) | \(=\) | \(2\) |
| \(y\) | \(=\) | \(2(x - 3)^2 - 8\) |
| \(\) | \(=\) | \(2(x^2 - 6x + 9) - 8\) |
| \(\) | \(=\) | \(\boldsymbol{2x^2 - 12x + 10}\) |
The axis is \(x = 3\). \((1,\ 0)\) is \(2\) units to its left, so its mirror image is \(2\) units to the right: the other \(x\)-intercept is \(x = \mathbf{5}\).
(i) Find where the line \(y = x + 1\) meets the parabola \(y = x^2 - 3x + 4\).
(ii) For which value of \(c\) does the line \(y = x + c\) touch the parabola at exactly one point?
(i) Set the two expressions for \(y\) equal:
| \(x^2 - 3x + 4\) | \(=\) | \(x + 1\) |
| \(x^2 - 4x + 3\) | \(=\) | \(0\) |
| \((x - 1)(x - 3)\) | \(=\) | \(0\) |
\(x = 1\) gives \(y = 2\), and \(x = 3\) gives \(y = 4\). They meet at \(\mathbf{(1,\ 2)}\) and \(\mathbf{(3,\ 4)}\).
(ii) Now \(x^2 - 3x + 4 = x + c\), so \(x^2 - 4x + 4 = c\), that is \((x - 2)^2 = c\). This has exactly one solution only when \(c = 0\).
So \(c = \mathbf{0}\): the line \(y = x\) is a tangent, touching at \((2,\ 2)\).
Common pitfalls
Frequently asked questions
How do you find the vertex of a parabola?
Find the axis of symmetry with \(x = -\dfrac{b}{2a}\), then substitute that \(x\)-value to get \(y\). Or complete the square to write \(y = a(x - h)^2 + k\); the vertex is \((h,\ k)\).
How do you know if a parabola is concave up or down?
Look at the coefficient of x squared. If it is positive the parabola is concave up with a minimum turning point; if it is negative it is concave down with a maximum turning point.
How do you find the axis of symmetry from the x-intercepts?
The axis of symmetry is halfway between the x-intercepts, so add them and divide by 2.
Can a parabola have no x-intercepts?
Yes. If it is concave up with its vertex above the x-axis, or concave down with its vertex below the x-axis, it never crosses the x-axis.
How do you find where a line meets a parabola?
Set the two expressions for y equal to each other, rearrange to a quadratic equal to zero and solve. Substitute each x-value into the line to get the y-values. A repeated solution means the line is a tangent.