A curve has the equation $y=4x^2-24x+30$. A line of gradient $m$ passes through the point $(0,-10)$.

1. Factorise out the coefficient of $x^2$ from the $x$ terms

$4x^2-24x+30 = 4(x^2-6x)+30$

2. Complete the square inside the bracket

$= 4\left[\left(x-\frac{6}{2}\right)^2 - \left(\frac{6}{2}\right)^2\right] + 30$
$= 4[(x-3)^2 - 9] + 30$

3. Expand and simplify to reach the final form

$= 4(x-3)^2 - 36 + 30$
$= 4(x-3)^2 - 6$

1. Use the completed square form from part (i) to identify the minimum point

From part (i), $y = 4(x-3)^2 - 6$.
Since $4(x-3)^2 \ge 0$ for all real $x$, the minimum value of $y$ is $-6$.
This occurs when $x - 3 = 0$, which gives $x = 3$.

2. State the coordinates of the minimum point

Therefore, the minimum point has coordinates $(3, -6)$ (shown).

1. Find the exact x-intercepts of the curve by setting $y = 0$.

$4(x-3)^2 - 6 = 0$
$(x-3)^2 = \frac{6}{4} = \frac{3}{2}$
$x - 3 = \pm\sqrt{\frac{3}{2}} = \pm\frac{\sqrt{6}}{2}$
$x = 3 \pm \frac{\sqrt{6}}{2}$

2. Determine the y-intercept and the new turning point after applying the modulus.

When $x = 0$, $y = |30| = 30$.
The original minimum point is $(3, -6)$.
Reflecting the negative portion in the x-axis, the new turning point is a local maximum at $(3, 6)$.

3. Sketch the modulus graph with all key features labelled.

Draw a W-shaped curve entirely above or on the x-axis.
Label the y-intercept at $(0, 30)$.
Label the x-intercepts at $(3 - \frac{\sqrt{6}}{2}, 0)$ and $(3 + \frac{\sqrt{6}}{2}, 0)$.
Label the local maximum turning point at $(3, 6)$.