Let $f(x) = ext{sin}^{-1}(x + 5)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2006 - Paper 1

To find the gradient, we first differentiate $f(x)$:

f'(x) = rac{1}{ ext{ extsqrt}(1 - (x + 5)^2)}

Evaluating at $x = -5$, we get:

f'(-5) = rac{1}{ ext{ extsqrt}(1 - 0)} = 1.

The graph of $y = f(x)$ will be an arc of the sine inverse function, starting at $x = -5$ where y = -rac{ ext{ extPi}}{2} and reaching $x = 5$ where y = rac{ ext{ extPi}}{2}. It should be symmetric about the line $y = 0$ within the range of $x$.

Starting from $(1 + x)^{n}$:

Differentiating gives us:

rac{d}{dx}[(1 + x)^{n}] = n(1 + x)^{n - 1}.

Also using the binomial expansion, we get:

(1 + x)^{n} = inom{n}{0} + inom{n}{1} x + inom{n}{2} x^{2} + ... + inom{n}{n} x^{n}.

After differentiation, equate the two results to demonstrate the required expression.

From the previous part, by substituting $x = 3$ into the differentiated result:

n3^{n - 1} = inom{n}{0} + inom{n}{1}3 + inom{n}{2}3^{2} + ... + inom{n}{n}3^{n}.

To find the coordinates of point $U$, evaluate the intersection of the chord $PR$ with the y-axis:

Setting $x = 0$ in the equation of $PR$ gives:

$y = -apr$

Thus, the coordinates of $U$ are $(0, -apr)$.

To show this, find the point of intersection of the tangents at $P$ and $Q$ by solving the equations of the tangents:

Setting them equal provides: $px - aq^{2} = p + rq - apr.$ Substituting the values will yield $T$ as $(a(p + q), aq)$.

To prove this, determine the slopes of line segments $TU$ and then check if the product of the slopes is -1, indicating perpendicularity:

If $m_{TU} m_{parabola} = -1$, then they are indeed perpendicular.