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

To find the gradient, we need to differentiate $f(x)$.

The derivative of $f(x)$ is:

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

At the point where $x = -5$:

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

Thus, the gradient at the point is 1.

To sketch the graph of $y = f(x)$, you need to plot the function within the domain $[-6, -4]$. Given the range of [-rac{ ext{ extpi}}{2}, rac{ ext{ extpi}}{2}], the graph will be a curve starting at the point $(-6, f(-6))$ and ending at the point $(-4, f(-4))$ with a smooth continuous characteristic, resembling an inverse sine curve.

Using the binomial theorem, we have:

(1 + x)^n = ext{ extsum_{k=0}^{n}} {n ackslash k} x^k

Differentiating with respect to $x$ gives:

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

Thus,

n(1 + x)^{n - 1} = ext{ extsum_{k=0}^{n}} k {n ackslash k} x^{k - 1}

From the previous result, we can deduce:

n3^{n - 1} = ext{ extsum_{k=1}^{n}} {n ackslash k} 3^{k - 1}

This leads us to the conclusion shown in the question, by substituting appropriate values.

To find the coordinates of point $U$, we know that the chord $PR$ is given by:

y = rac{1}{2}(p + r)x - apr

To find the intersection with the y-axis, set $x=0$:

$U = (0, -apr)$

Using the equations of the tangents and applying the point-slope form, we can derive that the coordinates will simplify to:

$T = (a(p + q), aq)$

This can be validated through substituting back into the tangent equations.

To show that line $TU$ is perpendicular to the parabola's axis, we must find the slopes of the lines $TU$ and the axis.

If $TU$ has a slope of $m_{TU}$ and the axis has a slope of 0, then if $m_{TU} imes 0 = -1$ holds, we confirm that they are indeed perpendicular.