Porisma

Sketch reciprocal graphs

Reciprocate outputs while retaining every excluded input.

Reciprocate outputs without extending the domain

For g(x)=1/f(x)g(x)=1/f(x), change each non-zero output y to 1/y1/y. Keep its input x unchanged. Before sketching, mark every input excluded from f and every zero of f: none of these inputs belongs to g.

Work in this order

  1. Mark excluded inputs and zeros before moving any points.
  2. Reciprocate the heights of labelled points. Keep the sign of each height.
  3. Check the limits on each side of an excluded input and at both infinite ends.
  4. Sketch each continuous branch, using the direction and turning-point rules below.

For example, (2,4)(2,4) becomes (2,1/4)(2,1/4) and (3,−2)(3,-2) becomes (3,−1/2)(3,-1/2). An original zero has no reciprocal point.

Reciprocal rules, one feature at a time

Point (x,y)(x,y) with y≠0y\ne0

(x,1/y)(x,1/y).

y-intercept (0,a)(0,a) with a≠0a\ne0

(0,1/a)(0,1/a).

Zero at x=bx=b

Exclude b; a vertical asymptote occurs when nearby non-zero f values tend to 0.

Vertical divergence at excluded x=cx=c

Reciprocal tends to 0+0^+ for f(x)→+∞f(x)\to+\infty and 0−0^- for f(x)→−∞f(x)\to-\infty, but c remains excluded; this is a missing limit point, not automatically an x-intercept.

Non-horizontal oblique asymptote y=dx+ey=dx+e

At the relevant infinite end, f grows in magnitude and the reciprocal tends to 0.

Finite non-zero horizontal limit L

Reciprocal horizontal limit 1/L1/L.

Non-zero local maximum (g,h)(g,h)

Local minimum (g,1/h)(g,1/h) on a continuous non-zero neighbourhood.

Non-zero local minimum (i,j)(i,j)

Local maximum (i,1/j)(i,1/j) on such a neighbourhood.

Positive/negative output

Same sign.

Increasing/decreasing on a continuous non-zero interval

Decreasing/increasing on that interval.

f(x)=±1f(x)=\pm1

The two graphs intersect at the same point.

Keep holes distinct from intercepts

The native source table calls zero limits at former vertical asymptotes “x-intercepts.” That printed statement remains in the review. On your sketch, do not fill excluded points: 1/f(x)1/f(x) cannot be zero at an input where f has a finite non-zero value, and simplifying a formula does not restore a missing input. Likewise, a zero of an arbitrary discontinuous f does not alone prove a nearby pole; inspect the limit.

The source sidebar’s invariance claim concerns non-degenerate translations/scalings: they preserve the turning-point count, while metric shape and the position of a symmetry axis need not stay fixed. Reciprocation can change the branch structure and the number/type of extrema.

Apply the rules to the examples

The calculator activity compares a line with root −2 and a quadratic with roots −1 and 3. Its four sketches include the Teacher labels.

Example 17 supplies three schematic graphs. Check these features in their reciprocals: the positive bowl has maximum (0,1/a)(0,1/a); the rational sketch has a hole at (0,0)(0,0); the oblique sketch has a hole at (−1,0)(-1,0) and extrema (−2,−1/4),(1/2,1/2)(-2,-1/4),(1/2,1/2). Do not infer an unknown numeric y-intercept from schematic geometry.

In Example 18, follow the labelled points A and B: A becomes the reciprocal maximum (3,−1)(3,-1), B gives the pole x=1x=1, the left horizontal asymptote becomes y=1/2y=1/2, and the original excluded origin stays a hole. None of these question sketches supplies a unique formula for f.

Try these examples

Open an example in chapter practice. Try it first, then read its solution. Use “Learn this step” to return to this unit.