1BMoreReciprocals


 * More Reciprocal Graphs **

When drawing the reciprocal of a graph, keep the following things in mind:
 * A vertical asymptote will occur anywhere the original graph has an x-intercept.
 * The reciprocal of 1 is 1 so anywhere y = 1 (or –1), the original graph and the reciprocal graph will intersect.
 * Anywhere 0 < y < 1, the reciprocal will be larger than 1.
 * Anywhere the y-value is larger than 1, the reciprocal will be between 0 and 1.
 * The reciprocal of a negative is negative, so the reciprocal graph will stay in the same quadrant of the axes as the original.
 * The x-axis will be a horizontal asymptote anywhere the original y-value increases to infinity.
 * Turning Points: The reciprocal graph will have a turning point at the same x-coordinate as a turning point on the original graph.

For a demonstration of this process in Powerpoint, see the 3rd slide of the following file (you may already have it):

Example

math \text {Sketch } y = \dfrac {1}{(x+2)^2 - 1} math by considering it as the reciprocal of y = (x+2) 2 – 1



The graph of y = (x+2) 2 – 1

x-intercepts at –1, –3

y-intercept at y = 3

Minimum at (–2, –1)

First draw in a vertical asymptote at each x-intercept and a horizontal asymptote along the x-axis.



Asymptotes drawn in with the equations x = –1, x = –3, y = 0

The top right section looks a bit like a straight line cutting the x-axis at –1, so the reciprocal in that section looks a bit like a section of a hyperbola. The original and reciprocal will intersect at y = 1. The same reasoning applies to the top left section.

Notice that the right section intersects the y-axis.

This will be at y =1/3 because the original has a y-intercept at y = 3.

To draw the central section, consider the following:
 * The center section is negative so the reciprocal will be negative.
 * The turning point is at x = –2 so the reciprocal will have a turning point at x = –2
 * In this case, the turning point is at y = –1 so the reciprocal turning point will also be at y = –1.
 * The center section is always between y = 0 and y = –1 so the reciprocal will always be below y = –1.
 * The original curve is at y = 0 when x = –1 and x = –3 so the reciprocal will approach negative infinity as x approaches those values.

**Final Graph:**

math y = \dfrac{1}{(x+2)^2 - 1} \quad \textit{(in red)} math

Asymptotes at x = –1, x = –3, y = 0

y-intercept at y = 0.33

local maximum at (–2, –1)

Domain: math x \in R \backslash \lbrace -1, -3 \rbrace math

Range: math \lbrace y \leqslant -1 \rbrace \cup \lbrace y > 0 \rbrace math

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