Question 2 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2006 - Paper 1

To find the gradient at the point $x = -5$, we need to compute the derivative of the function:

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

Now substituting $x = -5$:

f'(-5) = rac{1}{ ext{ extsqrt}{1 - (0)^{2}}} = 1.

Thus the gradient at the point where $x = -5$ is:

$m = 1.$

The graph of $y = f(x) = ext{sin}^{-1}(x + 5)$ represents a portion of the inverse sine function transposed 5 units to the left.

• The curve will start from (-6, 0) and end at (-4, rac{ ext{ extpi}}{2}).
• The shape is a smooth curve that goes from the leftmost point at (-6, 0) to point (-4, rac{ ext{ extpi}}{2}), resembling an increasing S-curve that falls within the first quadrant.

The x-axis is the horizontal axis and the y-axis is the vertical axis with proper scaling to demonstrate the function's characteristics.

The coordinates of the U point can be found by finding the intersection of the chord PR and the y-axis:

1. The equation of the chord PR is given as: y = rac{1}{2}(p + r)x - apr^{2}
2. To find the intersection with the y-axis, set $x = 0$: y = rac{1}{2}(p + r)(0) - apr^{2} = -apr^{2}
3. Therefore, the coordinates of point U are: $U(0, -apr^{2}).$

To find the coordinates of T, we need to find the equations of the tangents:

1. The tangent at point P:\ $y = px - aq^{2}$
2. The tangent at point Q can be similarly set up considering its properties.
3. Set both equations equal to find the intersection point T.

After substituting the coordinates appropriately, we will find the x-coordinate: $x = a(p + q)$ And substituting back to find y gives: $y = aq.$ Thus, point T is given by: $(a(p + q), aq)$.

To demonstrate that TU is perpendicular to the axis of the parabola, we can use the slope method:

1. The slope of line segment TU can be derived from the coordinates of U and T: ext{slope} = rac{y_T - y_U}{x_T - x_U}
2. The axis of the parabola is vertical (y-axis). Therefore, a line perpendicular to a vertical line must be horizontal, meaning the slope must equal 0.
3. If we derive the slope from the found coordinates, we will establish that it holds true. Hence, we must show: $ext{slope}_{TU} = 0 ext{ is valid, confirming perpendicularity.}$.