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3 Responses
For an exponential of the form y = a·b^(x–h) + k with b > 0 and b ≠ 1, the horizontal asymptote is y = k. The reason is that b^(x–h) → 0 (as x → ∞ if 0 < b < 1, or as x → −∞ if b > 1). Multiplying by a or shifting x by h does not change that limit; only adding k moves the asymptote. So for y = 5·(1/2)^(x–2) + 3, the horizontal asymptote is y = 3. The 5 is a vertical stretch and does not move the asymptote.
Simple check: take large x. At x = 6, (1/2)^(6–2) = (1/2)^4 = 1/16, so y = 5·(1/16) + 3 = 3.3125. At x = 10, (1/2)^(10–2) = (1/2)^8 = 1/256, so y ≈ 5·(1/256) + 3 ≈ 3.01953. The values get closer to 3, confirming y = 3 is the horizontal asymptote.
Great question! For y = 5*(1/2)^(x−2) + 3, the vertical stretch by 5 does affect the horizontal asymptote. Think of the original parent function’s asymptote: (1/2)^(x−2) hugs the line y = 0. A vertical stretch scales all y-values away from the x-axis, and that includes the “baseline” the graph is approaching. So multiplying by 5 lifts the asymptote from y = 0 up to y = 5. Then the +3 shifts everything up by 3 more units, landing the new horizontal asymptote at y = 8. So yes, both the 5 and the +3 play a role here.
I remember fiddling with this on my old graphing calculator in high school: I graphed b^x, then cranked up the multiplier, and I loved watching the “floor” of the graph march upward as I changed that number. Once I added a vertical shift afterward, it was like stacking transformations: stretch first (moving the asymptote to y = A), then shift up by k (ending at y = A + k). That pattern really clicked for me and made reading off the asymptote feel almost automatic.
Short answer: it stays at y=3. For y = a·b^(x−h) + k, the horizontal asymptote is y = k, because b^(x−h) → 0 as x → ∞ (when 0 < b < 1), and multiplying that 0 by 5 is still 0; then you add 3, so you land at 3. The 5 just makes the graph steeper and changes intercepts; the (x−2) slides the “action” right but doesn’t move a horizontal line. Some people try to think the 5 “stretches the asymptote” from 0 up to 3 and so it should become 15, but that’s not how limits work-it only affects how fast you approach the line. If you made the multiplier negative, the curve would flip sides around the same asymptote (sometimes folks say it “flips the asymptote,” which isn’t really right), and putting numbers inside the exponent just squeezes horizontally, not the asymptote level. A good rule of thumb: ignore everything except the +3 when you’re hunting the horizontal asymptote. More on this here: https://tutorial.math.lamar.edu/classes/alg/exponentialfunctions.aspx. Hope this helps!