Hi thanks for the answers and I got most of them right but I got a question.
On Q3 what does the graph suppose to look like?
And on Q7a how did you get +4 on 8x^-1 to get the value for c. because if x = 2 I got 1/8(2) so I got 1/16 because I thought the power of -1 meant 1 over.
horizontal asymptotes are lines of the form y = c
so if you can find x and y values that are not possible for the function to attain, you found asymptotes.
think of 1/x, are there any restrictions on x? you bet there are, x cannot be zero, or we are undefined, so guess what? x = 0 is a vertical asymptote, our graph cannot reach it.
but this also affects y. because 1/x can never be zero with the restrictions we have on x. so y cannot be zero no matter what x we plug in, so y = 0 is a horizontal asymptote