����ý

If a quadratic equation can be factorised, the factors can be used to find where the graph crosses the \(x\)-axis.

Sketch \(y = x^2 + x − 6\)

The quadratic factorises to give \((x + 3)(x − 2)\) so the solutions of the equation \(x^2 + x − 6 = 0\) are \(x = -3\) and \(x = 2\).

The graph of \(y = x^2 + x − 6 \) crosses the \(x\)-axis at \(x = -3\) and \(x = 2\).

The coefficient of \(x^2\) is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

To find where the graph crosses the \(y\)-axis, work out \(y\) when \(x\) = 0. So:

\(0^2 + 0 – 6 = -6\) so the graph crosses the \(y\)-axis at \(y = -6\).

Sketch \(y = x^2 − 6x + 9\)

The quadratic factorises to give \((x − 3)(x − 3)\) so the only solution of the equation \(x^2 − 6x + 9\) = 0 is \(x = 3\).

The graph of \(y = x^2 − 6x + 9\) touches the \(x\)-axis at \(x = 3\).

The coefficient of \(x^2\) is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

To find where the graph crosses the \(y\)-axis, work out \(y\) when \(x = 0\):

\(0^2 – 6 \times 0 + 9 = 9\) so the graph crosses the \(y\)-axis at \(y = 9\).

Sketch \(y = x^2 – 6x + 4\).

The coefficient of \(x^2\) is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

Writing \(y = x^2 – 6x + 4\) in completed square form gives \(y = (x – 3)^2 – 5\).

Squaring positive or negative numbers always gives a positive value. The lowest value given by a squared term is 0, which means that the minimum value of the term \((x – 3)^2 – 5\) is given when \(x = 3\). This also gives the equation of the line of symmetry for the quadratic graph.

The value of \(y\) when \(x = 3\) is -5. This value is always the same as the constant term in the completed square form of the equation.

So the graph of \(y = x^2 – 6x + 4\) has a line of symmetry with equation \(x = 3\) and a minimum turning point at (3, -5).

When \(x = 0\), \(y = 4\). So the graph crosses the \(y\)-axis at (0, 4).