The sketch below shows the graph of $$f(x)=\frac{6}{x-4} + 3$$ The asymptotes of $f$ intersect at A - NSC Mathematics - Question 4 - 2017 - Paper 1

To find the coordinates of B, we set the output of the function to zero to find the $x$-intercept:

$0 = \frac{6}{x-4} + 3$

Solving for $x$ gives:

$\frac{6}{x-4} = -3$

$6 = -3(x-4)$

Expanding and solving:

$6 = -3x + 12$

$3x = 12 - 6$

$3x = 6$

$x = 2$

Thus, the coordinates of B are B(0, 2).

To find the coordinates of C, we set $x$ to zero to find the $y$-intercept:

$y = \frac{6}{0 - 4} + 3$

Calculating gives:

$y = \frac{6}{-4} + 3 = -\frac{3}{2} + 3 = \frac{3}{2}$

Thus, the coordinates of C are C(0, 0).

The average gradient between points B and C is given by the formula:

$\text{Average Gradient} = \frac{f(y_C) - f(y_B)}{x_C - x_B}$

At B, the coordinates are (0, 2) and at C, the coordinates are (2, 0):

$\text{Average Gradient} = \frac{0 - 2}{2 - 0} = \frac{-2}{2} = -1$

Thus, the average gradient is -1.

The line of symmetry can be determined by considering the function and its behavior. The symmetry can be expressed in the form of the line:

$y = mx + b$

For this function, we can assume a line of the form $y = -x + 7$ with a slope of -1, which will intersect at $y$ with a positive intercept. This assumes symmetry around the vertical asymptote at x = 4, and the line would reflect across this axis.