Quadratic Graphs
Master quadratic graphs for NSW Year 12 Mathematics Standard 2. In this topic you recognise a quadratic relationship across its equation, table and graph, and sketch the parabola \(y=ax^{2}+bx+c\) — reading its vertex, axis of symmetry and \(x\)- and \(y\)-intercepts.
You will learn to find the axis of symmetry from the midpoint of the x-intercepts, work out the vertex by substituting back, and interpret parabolic models — the greatest height of a thrown ball or a water jet and the valid domain — a core Standard 2 skill for solving practical problems.
Theory
A quadratic relationship has an \(x^{2}\) term as its highest power, \(y=ax^{2}+bx+c\), and graphs as a parabola. This Year 12 Standard 2 (NSW) guide shows how to recognise a quadratic across an equation, table or graph, find the vertex, axis of symmetry and the x- and y-intercepts, use the midpoint of the x-intercepts, and interpret parabolic models.
A quadratic relationship has an \(x^{2}\) term as its highest power, so its equation has the form \(y=ax^{2}+bx+c\) with \(a\neq 0\). Its graph is a parabola — a smooth, symmetric curve that is U-shaped when \(a>0\) and ∩-shaped (upside-down) when \(a<0\).
Every parabola has a turning point called the vertex (the lowest point when it opens up, the highest point when it opens down) and a axis of symmetry — the vertical line through the vertex that makes the two sides mirror images. You can recognise a quadratic across representations: an \(x^{2}\) equation, a U-shaped graph, or a table whose second differences are constant.
The graph cuts the axes at its intercepts. The \(y\)-intercept is where \(x=0\), giving \(y=c\); the \(x\)-intercepts are where \(y=0\). A key Standard 2 shortcut: the axis of symmetry sits exactly halfway between the two \(x\)-intercepts, and substituting that \(x\) back into the equation gives the vertex.
A quadratic relationship is written in general form:
The \(y\)-intercept is the constant term (set \(x=0\)):
The axis of symmetry is the midpoint of the two \(x\)-intercepts \(x_1,x_2\) (equivalently \(x=-\dfrac{b}{2a}\)):
How to sketch and read a parabola
- Confirm it is quadratic. The highest power is \(x^{2}\); in a table the second differences are constant.
- Direction. \(a>0\) opens upward (a minimum); \(a<0\) opens downward (a maximum).
- \(y\)-intercept. Set \(x=0\): the curve cuts the \(y\)-axis at \((0,c)\).
- \(x\)-intercepts. Set \(y=0\) (or read them off the graph) — where the curve cuts the \(x\)-axis.
- Axis of symmetry. Take the midpoint of the \(x\)-intercepts, \(x=\dfrac{x_1+x_2}{2}\).
- Vertex. Substitute that \(x\) back into the equation to get the vertex \(y\); state the vertex as \((x,y)\).
| \(x\) | \(-2\) | \(-1\) | 0 | 1 | 2 |
|---|---|---|---|---|---|
| \(y\) | 1 | \(-2\) | \(-3\) | \(-2\) | 1 |
A quadratic has constant second differences and a symmetric set of values.
| \(\text{first differences}\) | \(:\) | \(-3,\,-1,\,+1,\,+3\) |
| \(\text{second differences}\) | \(:\) | \(+2,\,+2,\,+2\ \text{(constant)}\) |
| \(\Rightarrow\) | \(\) | \(\text{quadratic}\) |
| \(\text{lowest } y=-3 \text{ at } x=0\) | \(\Rightarrow\) | \(\text{vertex }(0,-3)\) |
Constant second differences confirm a quadratic; the vertex is \((0,-3)\).
Read where the curve cuts each axis, then use the midpoint for the vertex.
| \(x\text{-intercepts}\) | \(:\) | \((-3,0),\ (1,0)\) |
| \(y\text{-intercept}\) | \(:\) | \((0,-3)\) |
| \(\text{axis } x\) | \(=\) | \(\dfrac{-3+1}{2}=-1\) |
| \(y\) | \(=\) | \((-1)^2+2(-1)-3=-4\) |
\(x\)-intercepts \((-3,0)\) and \((1,0)\); \(y\)-intercept \((0,-3)\); vertex \((-1,-4)\).
The axis is halfway between the \(x\)-intercepts; substitute for the vertex.
| \(\text{axis } x\) | \(=\) | \(\dfrac{-1+5}{2}=2\) |
| \(y\) | \(=\) | \((2)^2-4(2)-5\) |
| \(\) | \(=\) | \(4-8-5=-9\) |
| \(\therefore\ \text{vertex}\) | \(=\) | \((2,-9)\) |
Axis of symmetry \(x=2\); vertex \((2,-9)\).
The maximum is at the vertex, midway between the two ground times.
| \(t\) | \(=\) | \(\dfrac{0+8}{2}=4\) |
| \(h\) | \(=\) | \(40(4)-5(4)^2\) |
| \(\) | \(=\) | \(160-80=80\) |
| \(\text{valid domain}\) | \(:\) | \(0 \le t \le 8\) |
Maximum height \(80\) metres at \(t=4\) s; the model is valid for \(0 \le t \le 8\) (while the ball is in the air).
Common pitfalls
Frequently asked questions
What shape is the graph of a quadratic?
It is a parabola: a smooth, symmetric curve. When the coefficient of x squared is positive the parabola opens upward (U-shaped) and has a lowest point; when it is negative the parabola opens downward and has a highest point.
How do you find the vertex of a parabola?
Find the axis of symmetry first, which is the midpoint of the two x-intercepts, x equals (x1 plus x2) divided by 2 (or x equals minus b over 2a). Then substitute that x-value back into y equals ax squared plus bx plus c to get the vertex y-value. State the vertex as the point (x, y).
How do you find the axis of symmetry from the x-intercepts?
Take the midpoint of the two x-intercepts. If the parabola cuts the x-axis at x1 and x2, the axis of symmetry is the vertical line x equals (x1 plus x2) divided by 2. This works because a parabola is symmetric, so its turning point sits exactly halfway between the intercepts.
What are the x- and y-intercepts of a parabola?
The y-intercept is where the curve cuts the y-axis, found by setting x equal to 0, which gives y equals c (the constant term). The x-intercepts are where the curve cuts the x-axis, found by setting y equal to 0. A parabola can have two, one, or no x-intercepts.
How do you know if a parabola opens up or down?
Look at the sign of a, the coefficient of x squared. If a is positive the parabola opens upward and the vertex is a minimum. If a is negative it opens downward and the vertex is a maximum.
How do you recognise a quadratic from a table of values?
Work out the differences between successive y-values, then the differences of those differences. If these second differences are constant (and not zero), the table represents a quadratic relationship. The y-values will also be symmetric about the vertex.