Given: $H(x) = \frac{-3}{x-1} + 2$ 4.1 Write down the equations of the asymptotes of $h$ - NSC Mathematics - Question 4 - 2020 - Paper 1

The axis of symmetry for a rational function can be found by analyzing the behavior of the function. In this case:

• The gradient can be deduced by observing that it generally runs through the midpoint of the asymptotic behavior.
• For this function, the formula for the axis of symmetry is ( y = -x + c ). To find the exact value of ( c ), we substitute a point where the function intersects the axes, yielding:
  [ H(1) = 2 \Rightarrow \text{Axis: } y = -x + 3 ]

To sketch the graph of ( H(x) ):

• Plot the vertical asymptote at ( x = 1 ) and the horizontal asymptote at ( y = 2 ).
• The function will approach these lines but never touch them.
• Calculate intercepts by setting ( H(x) = 0 ) to find the x-intercept and substituting ( x = 0 ) for the y-intercept.
• The intercept calculations yield the x-intercept and the y-intercept values which are critical for accurate representation.

This creates an 'A' shaped graph that meets the axes at calculated points while approaching the asymptotes.

The coordinates of point A, the turning point of the function ( f ), are calculated by finding the vertex of the parabola represented by ( f(x) = \frac{1}{2}(x + 5)^2 - 8 ). Using the vertex formula, we find that:

• The x-coordinate of the vertex is ( x = -5 ), and substituting this back into the function gives the y-coordinate:
  [ f(-5) = \frac{1}{2}(0) - 8 = -8 ]
  Thus, the coordinates are ( A(-5, -8) ).

The range of the function ( f ) can be determined from the vertex form of the quadratic. Since the parabola opens upwards and the vertex represents the minimum point:

• The minimum value of ( f ) is ( -8 ). Therefore, the range is ( [-8, \infty) ).

We calculate the coordinates where ( f ) intersects ( g ). The coordinates of intersection are represented as ( D(m; n) ). By substituting the respective equations into each other, we determine:

• For the function, we find ( g(-5) = 1 ) leading to values of ( m = -5 ) and ( n = 2 ).

To find the area of trapezoid OCDE, we calculate the area by the formula:
[ \text{Area} = \frac{1}{2}(b_1 + b_2)h ]

• Bases: ( OE = 5 ) and the y-coordinate difference at point C, which is determined to be ( 2 ).
• Substituting values results in the area calculation:
  [ \text{Area} = \frac{1}{2} \times (10 + 5) \times 5 = 10 ]

To find the inverse of the function ( g(x) = \frac{1}{2}x + \frac{9}{2} ), we swap x and y and solve for y:

1. Set ( x = \frac{1}{2}y + \frac{9}{2} )
2. Rearranging gives: [ y = 2x - 9 ]
   Thus, the equation of the inverse is ( g^{-1}(x) = 2x - 9 ).

To determine the point of contact, we equate ( H(x) ) with the function ( f ), representing a tangent point:

1. Substitute ( g^{-1}(c) + k ) into the function and solve ( H(x) = f(x) ).
2. Calculate the derivatives to establish common tangent characteristics and deduce values based on previously established coordinates, leading to the required point of contact.