Week 13: Graphing Reciprocals of Linear and Quadratic Functions

Published by on May 13, 2018 | Leave a response

Reciprocal is the quantity obtained by dividing 1 by a given quantity other than 0. The reciprocal of f(x),f(x)\neq 0, is \frac{1}{f(x)}.

The graphing of the reciprocals is a hyperbola.

*In order to graph the reciprocal of a function, we need to find some information:

– Asymptote: the invisible line that the hyperbola slowly reaches without ever touching. In other words, the asymptote is the restriction for the hyperbola.
+ Horizontal asymptote: f^{-1}(x) = \frac{1}{f(x)}, f(x)\neq 0. With every value of f(x), we will never get f^{-1}(x) =0. That’s why we have the horizontal asymptote for y= \frac{1}{f(x)} as y=0.
+ Vertical asymptote: with \frac{1}{f(x)}, f(x) must be different from 0. That’s why the vertical asymptote will be the value that makes f(x) = 0, which has the same x value of the x-intercept.

– Invariant points: the intersections of the linear/quadratic function and its reciprocal function. Since the reciprocal of 1 is 1, of -1 is -1, the points which has y=1 or y=-1 are the only points that the line/parabola shares with the corresponding hyperbola.

*Based on this information, we can now graph a reciprocal of a linear/quadratic function with this general guidance:

  • Draw the original f(x) function (f^{-1}(x) = \frac{1}{f(x)})
  • Identify the asymptotes.
  • Find the invariant points
  • The asymptotes create sections on the xy-plane. The hyperbola only exists in the sections which the original function goes through. In each of those sections, draw a curve which goes through the invariant point(s) and gets closer to the asymptotes without touching them.

 

*Reciprocal of a linear function: y = \frac{1}{4x+3}
– Draw the original linear function: 4x+3

– Identify asymptotes, which are, in this case, vertical: x=\frac{-3}{4}, horizontal: y=0

– Find the invariant points: (-0.5,1) and (-1,-1)

– Draw the curves/hyperbola.

*Reciprocal of a quadratic function:

    • 2 roots: y=\frac{1}{x^2 -5x-14}

– Draw the original parabola:

– Identify the asymptotes

– Find the invariant points

– Draw the curves/hyperbola:

    • 1 root: 3(x-4)^2

-Draw the original parabola:

– Identify the asymptote:

– Find the invariant points:

– Draw the curves/ hyperbola:

    • ***No root: -2x^2 -1

This is a special case since there is no x-intercept -> no vertical asymptote. In order to graph this reciprocal, we find the maximum/minimum point of the hyperbola, which is the reciprocal of the y value of the vetex.
– Draw the original parabola:

– The vertex is (0,-1) so the minimum point of the hyperbola will be (0, -1)

– Draw the hyperbola:

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