Important question on graphing

anpanman SG

Let's say we have to draw the graph of y = 1 / f(x) given the graph of y = f(x). First we have to determine the vertical asymptotes of the new graph, which is where the original graph cuts the x-axis. But what happens if the original graph has its own vertical asymptotes? Do we include the original vert. asymptotes in the new graph or just use the new vertical asymptotes?

And is drawing the graph of y^2 = -f(x) the same as drawing y = - (f(x))^0.5

wee_ws SG

Hi,

Good questions that reflect deep thinking :)

If y = f(x) has a vertical asymptote at x = a, then y = 1/f(x) will have an x-intercept at x = a. For example, let f(x) = 1/(x - a).

To obtain the graph of y^2 = -f(x) from the graph of y = f(x), two actions are involved:

1. Reflection about the x-axis.

2. y^2-transform.

As for y = - (f(x))^0.5, it is the lower half of y^2 = f(x), and y^2 = f(x) is not y^2 = -f(x) to begin with.

If you are unable to grasp fluffy abstract concepts in maths, always use specific examples to convince yourself :) In this case, let f(x) = x. Use a GC to graph and compare:

y = +/- sqrt(-x) and y = -sqrt(x).

Thanks!

Cheers,
Wen Shih