I’m trying to get confident sketching reciprocal graphs with transformations, and I keep second-guessing where the two branches should go and whether intercepts exist.
Example 1: y = -2/(x+1) + 3. I think the vertical asymptote is x = -1 and the horizontal asymptote is y = 3. For intercepts, I set y = 0 and got x = -1/3, and setting x = 0 gives y = 1. That seems fine. But when I try to place the branches, I’m unsure which side of x = -1 each branch sits on and whether they approach y = 3 from above or below. I tried to reason “-2/(x+1)” is like -1/x reflected and stretched, then shifted up by 3, but my sketch keeps putting the point (0,1) on a branch that doesn’t match the asymptote behavior I expect.
Example 2: y = 1/(2x – 4). I rewrote it as y = 1/[2(x – 2)]. I’m saying vertical asymptote x = 2 and horizontal asymptote y = 0. Do I need to think about the 2 as a horizontal scaling that affects how “tight” the curve is, or is that irrelevant for a quick sketch? Also, do these branches stay in the same quadrants relative to the asymptotes as y = 1/(x – 2), or does the factor 2 flip anything? I don’t think it flips, but I’m not 100% sure.
What’s the most reliable, minimal set of checks to place the two branches correctly after translations and a possible negative coefficient? Is there a simple sign test per region (left/right of the vertical asymptote) that avoids mistakes, or a quick rule I can apply so I stop misplacing points and branches?
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3 Responses
Use this quick rule: for y = a/[b(x − h)] + k, the asymptotes are x = h and y = k, and you place branches by a sign test on each side of x = h-evaluate a/[b(x − h)] at one x < h and one x > h; if it’s positive the branch lies above y = k (approaches from above), if negative it lies below (approaches from below); the factor b only stretches horizontally (no flip), and only the sign of a decides which side is above/below. Example: y = −2/(x+1) + 3 → asymptotes x = −1, y = 3; test x = −2 gives −2/(−1) = +2 so the left branch is above y = 3, test x = 0 gives −2/1 = −2 so the right branch is below y = 3; intercepts are x = −1/3, y = 1; for y = 1/(2x − 4) = 1/[2(x − 2)], asymptotes x = 2, y = 0, left test x = 1 gives negative (below), right test x = 3 gives positive (above)-same quadrants as 1/(x − 2); see: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/graphs-of-rational-functions/a/transformations-of-rational-functions.
My go-to “don’t-get-lost” trick is to rewrite any reciprocal in the form y = d + k/(x − h). Then the vertical asymptote is x = h and the horizontal asymptote is y = d. The branch placement is a pure sign test: y − d = k/(x − h). So to the right of the vertical asymptote (x > h), the sign of y − d matches the sign of k; to the left (x < h), it’s the opposite. In other words: if k > 0, the right-hand branch sits above y = d and the left-hand branch sits below; if k < 0, swap those. That’s the whole placement rule, and I still like to sanity-check with one easy point on each side (I’ve definitely been tripped up by a stray minus sign before coffee). As a bonus: intercepts are quick-y-intercept is y(0) if 0 ≠ h, and x-intercept exists only if d ≠ 0 (since k/(x − h) = −d must be solvable). A nice recap lives here: https://www.khanacademy.org/math/algebra2/rational-expressions-equations/rational-functions/a/graphs-of-rational-functions-intro
Example 1: y = −2/(x + 1) + 3. That’s d = 3, h = −1, and k = −2. Asymptotes x = −1 and y = 3 are right. Since k < 0, the right side of x = −1 lies below y = 3, and the left side lies above y = 3. Your intercepts check out: y(0) = 1 (which is indeed below y = 3 and to the right of x = −1), and setting y = 0 gives x = −1/3 (also to the right of −1). Near x = −1 from the right, the function dives to −∞; from the left, it shoots to +∞; and both branches approach y = 3 (below on the right, above on the left). Example 2: y = 1/(2x − 4) = (1/2)·1/(x − 2). Here d = 0, h = 2, k = 1/2 > 0, so the right branch sits above the x-axis and the left branch sits below. The factor 2 does not flip anything; it just scales the “height” of the curve (you can think of it as making the graph a bit less steep), which is irrelevant for quick placement. Because d = 0 and the numerator is nonzero, there’s no x-intercept; the y-intercept is y(0) = −1/4, which lands on the left/below branch as expected.
I’d use a small, mechanical checklist. First find the asymptotes: for y = k/(x − h) + v the vertical is x = h and the horizontal is y = v; more generally for y = k/(ax + b) + d you get x = −b/a and y = d (or y = 0 when d = 0). Then do a sign test on each side of the vertical asymptote: evaluate the reciprocal part k/(…) at one x just left and one just right; if it’s positive the branch sits above y = v there, if negative it sits below. Example 1: y = −2/(x + 1) + 3 has x = −1, y = 3, with x-intercept x = −1/3 and y-intercept y = 1. Test x = −2: −2/(−1) + 3 = 5, so the left branch is above 3; test x = 0: −2/1 + 3 = 1, so the right branch is below 3. Hence it’s the “upper-left” and “lower-right” pair; you get y → +∞ as x → −1−, y → −∞ as x → −1+, and it approaches y = 3 from above on the left and from below on the right. Example 2: y = 1/(2x − 4) = 1/[2(x − 2)] has x = 2 and y = 0; no x-intercept, and y-intercept −1/4. The factor 2 just tightens the curve horizontally; a quick test x = 3 gives 1/2 > 0 (right branch above the axis) and x = 1 gives −1/2 < 0 (left branch below), so it’s the same quadrants as y = 1/(x − 2). As a quick rule, a negative coefficient flips the branches to the opposite pair of quadrants relative to the center (h, v); a positive one keeps them. I’m fairly sure this covers what you need; I might be oversimplifying the role of the linear factor in the denominator a bit, but for quick sketches the two sign checks plus the intercepts are usually enough to place everything correctly.