Express each of $4x^2 - 6x - 1$ and $-x^2 - 10x + 3$ in the form $a(x+b)^2 + c$, where $a$, $b$ and $c$ are constants. Using your answers in (a), show that the curves $y = 4x^2 - 6x - 1$ and $y = -x^2 - 10x + 3$ will intersect.

1. Complete the square for the first expression

$4x^2 - 6x - 1$
$= 4\left(x^2 - \frac{3}{2}x\right) - 1$
$= 4\left[\left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right] - 1$
$= 4\left(x - \frac{3}{4}\right)^2 - 4\left(\frac{9}{16}\right) - 1$
$= 4\left(x - \frac{3}{4}\right)^2 - \frac{9}{4} - \frac{4}{4}$
$= 4\left(x - \frac{3}{4}\right)^2 - \frac{13}{4}$

2. Complete the square for the second expression

$-x^2 - 10x + 3$
$= -\left(x^2 + 10x\right) + 3$
$= -\left[(x + 5)^2 - 5^2\right] + 3$
$= -(x + 5)^2 + 25 + 3$
$= -(x + 5)^2 + 28$

1. State the completed square forms from part (a)

From (a),
$4x^2 - 6x - 1 = 4\left(x - \frac{3}{4}\right)^2 - \frac{13}{4}$
$-x^2 - 10x + 3 = -\left(x + 5\right)^2 + 28$

2. Identify the minimum and maximum values of the two curves

For $y = 4\left(x - \frac{3}{4}\right)^2 - \frac{13}{4}$, the minimum value is $-\frac{13}{4}$ (since $4 > 0$).
For $y = -\left(x + 5\right)^2 + 28$, the maximum value is $28$ (since $-1 < 0$).

3. Compare the extreme values to deduce intersection

Since the minimum value of the first curve ($-\frac{13}{4}$) is less than the maximum value of the second curve ($28$), and the first curve opens upwards while the second opens downwards, their ranges overlap. Since they are continuous curves, they must intersect.