����ý

\(x^2 + x - 6 = 0 \)

The equation factorises to give \((x + 3) (x - 2) = 0 \) so the solutions to 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 \(x = 2\).

\(x^2 - 6x + 9 = 0\)

The equation factorises to give \((x – 3)(x – 3) = 0\) so there is just one solution to the equation, \( x = 3\).

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

\(x^2 + 2x + 5 = 0\)

Using the quadratic formula to try to solve this equation, \(a = 1\), \(b = 2\) and \(c = 5\) which gives:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{-2\pm \sqrt{-16}}{2 \times 1}\)

The graph of \(y = x^2 + 2x + 5\) does not cross or touch the \(x\)-axis so the equation \(x^2 + 2x + 5 = 0\) has no roots.

The graph of the quadratic equation \(y = ax^2 + bx + c\) crosses the \(y\)-axis at the point \((0, c)\). The \(x\)-coordinate of any point on the \(y\)-axis has the value of 0 and substituting \(x = 0\) into the equation \(y = ax^2 + bx + c\) gives \(y = c\).

Find the \(y\)-intercept of the following quadratic functions: