ACARA v9 CONTENT DESCRIPTION “identify and graph quadratic functions, solve quadratic equations graphically and numerically, and solve monic quadratic equations with integer roots algebraically, using graphing software and digital tools as appropriate”
Builds on: Algebraic Expressions and Quadratic Factorisation (AC9M9A02). This unit builds directly on factorising monic quadratics and on plotting points in the Cartesian plane. Holding the graphical, numerical and algebraic views together is what makes quadratics feel like one idea rather than three disconnected procedures, and it prepares the modelling work that follows in Year 9.
What a quadratic function looks like
A quadratic function is any rule of the form y equals x squared plus b x plus c, and its graph is always a smooth U-shaped curve called a parabola. The squared term is what bends the line into a curve: as you move away from the centre in either direction, the output grows ever faster. This unit is about reading that curve, drawing it from a function, and solving the equation that asks where the curve meets the x-axis, using three complementary tools.
The basic parabola from a table
The cleanest way to meet a parabola is to build a table of values for the simplest case, y equals x squared. Choosing x from minus three to three gives outputs nine, four, one, zero, one, four, nine. Two features jump out immediately. The curve is symmetric: the value at minus two equals the value at two, because squaring erases the sign. And it has a lowest point, the vertex, sitting at the origin, from which both arms sweep upward. Because the coefficient of x squared is positive, the parabola opens upward and the vertex is a minimum.
y = x² from a table
Build the simplest quadratic point by point, then watch the points become a smooth U.
x
−3
−2
−1
0
1
2
3
y
9
4
1
0
1
4
9
Seven points from (−3, 9) to (3, 9) lie on y = x². Reveal the smooth curve through them.
Moving the parabola: axis and vertex
Shifting to a general monic quadratic keeps the same shape but moves it around the plane. For y equals x squared minus two x minus three, a table from minus two to four gives five, zero, minus three, minus four, minus three, zero, five. The same mirror symmetry appears, this time about the vertical line x equals one, which is the axis of symmetry. The vertex sits on that axis at the point one comma minus four, the lowest point of this particular curve. Plotting these points and joining them with a smooth curve, never a series of straight segments, produces the graph.
Symmetry and the vertex
A monic quadratic keeps the U-shape but slides around the plane; its mirror line passes through the vertex.
The parabola y = x² − 2x − 3. Reveal its axis of symmetry and its vertex.
Solving by reading the graph
Solving a quadratic equation means finding the x-values where y is zero, which graphically are the points where the parabola crosses the x-axis. For y equals x squared minus two x minus three, the curve cuts the axis at x equals minus one and x equals three; these crossing points are the solutions, also called the roots. Reading solutions straight off a graph is the graphical method, and it works for any quadratic, though it gives only as much precision as the graph allows.
Roots as x-axis crossings
Solving the equation means finding where y is zero, which is exactly where the curve crosses the x-axis.
Where does y = x² − 2x − 3 meet the x-axis? Reveal the roots you can read straight off the graph.
Solving monic equations by factorising
When a quadratic is monic and its roots are whole numbers, there is a fast exact method: factorising. To solve x squared minus five x plus six equals zero, factorise the left side into x minus two times x minus three. A product is zero only when one of its factors is zero, so x equals two or x equals three. These match the points where the corresponding parabola would cross the axis. Notice the roots are two and three, not minus two and minus three; the factors x minus two and x minus three are zero at positive values. This algebraic route is reserved for monic quadratics with integer roots, where the factors are easy to find.
Factorising to exact roots
When a monic quadratic has whole-number roots, factorising reads them off exactly with no graph needed.
Solve x² − 5x + 6 = 0 by factorising. Reveal the factors and the matching crossings.
When roots are not whole numbers
Not every quadratic cooperates so neatly. The equation x squared equals five has solutions x equals plus or minus the square root of five, which is irrational, approximately two point two four but never exactly that. Here factorising into integers is impossible, so you fall back on the graphical or numerical methods, reading the crossings from a graph or letting digital tools home in on the value step by step. Graphing software is invaluable for exactly these cases, and for quickly checking any solution you found by hand.
An irrational-root case
Not every quadratic factorises over the integers; when it does not, the graph and digital tools take over.
The equation x² = 5 has no integer factors. Reveal where its parabola crosses the x-axis.
Two cautions: no crossings, and which way it opens
Two cautions round out the picture. First, a parabola need not cross the x-axis at all; y equals x squared plus one sits entirely above the axis, so the equation x squared plus one equals zero has no real solutions, and its graph confirms this at a glance. Second, always confirm whether the curve opens up or down before trusting a vertex to be a minimum or a maximum. For the monic functions in this unit the x squared coefficient is positive, so every parabola opens upward and every vertex is a lowest point.
No crossings
A parabola need not meet the x-axis at all; when it does not, the equation has no real solutions.
Does y = x² + 1 ever reach the x-axis? Reveal the answer in the shaded gap below it.
Quick self-check
1. For y = x², what are the coordinates of the vertex?
2. Solve x² − 5x + 6 = 0.
3. The graph of y = x² − 2x − 3 crosses the x-axis at which points?