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3 Responses
Great question. I always double‑check this kind of thing too, because the “shift inside the exponent” can feel like it might tug the whole graph (and its asymptote) sideways.
Step 1: Start from the base function
– For y = 2^x, we know 2^x → 0 as x → −∞. So the horizontal asymptote is y = 0.
– This is because exponential decay toward 0 happens on the left side for bases bigger than 1.
Step 2: Apply a vertical shift
– For y = 2^x − 3, subtracting 3 just drops every y‑value down by 3.
– So the horizontal asymptote moves from y = 0 to y = −3.
– Quick limit check: lim_{x→−∞} (2^x − 3) = 0 − 3 = −3.
Step 3: Apply a horizontal shift inside the exponent
– For y = 2^{x−3} − 1, it looks like a right shift by 3 units. The natural worry is: does that move the horizontal asymptote?
– Two quick ways to see what happens:
Method A (substitute): Let t = x − 3. As x → −∞, we also have t → −∞. Then 2^{x−3} = 2^t → 0, so y = 2^{x−3} − 1 → 0 − 1 = −1. That gives asymptote y = −1.
Method B (rewrite): 2^{x−3} = 2^x / 8. Then as x → −∞, 2^x → 0, so 2^{x−3} → 0/8 = 0, and again y → −1.
– So the asymptote is y = −1. In other words, the horizontal shift did not change the y‑value of the horizontal asymptote; the vertical shift (the “−1”) did.
A handy rule of thumb
– Horizontal shifts f(x − h) do not change a horizontal asymptote y = L, because as x → ±∞, x − h → ±∞ as well, so the limiting y‑value stays L.
– Vertical shifts f(x) + k do change it: y = L becomes y = L + k.
– A small subtlety that sometimes makes me hesitate: for exponentials, 2^{x−3} = (1/8)·2^x. That’s a vertical scaling by 1/8, which in general would scale a nonzero asymptote L to (1/8)L. But here L = 0 for 2^x, and (1/8)·0 is still 0, so the asymptote doesn’t move. Then the “−1” moves it to y = −1. I sometimes momentarily mix up “scaling vs shifting,” but the limit settles it cleanly.
Simple worked example (to cement the idea)
– Consider y = 3^{x+2} + 5.
– Rewrite: 3^{x+2} = 9·3^x. Since 3^x → 0 as x → −∞, we have 9·3^x → 0.
– Then y → 0 + 5 = 5. So the horizontal asymptote is y = 5.
– Even though there’s a +2 inside the exponent (a left shift), the asymptote’s y‑value is dictated by the vertical +5.
Answering your question directly
– Yes: for these exponentials, only the vertical shift changes the horizontal asymptote’s y‑value.
– Quickest way to see it without plotting:
1) Note 2^{x−3} = 2^x/8, so the “toward 0” behavior stays “toward 0.”
2) Then apply the vertical shift at the end.
Your conclusions y = −3 for y = 2^x − 3 and y = −1 for y = 2^{x−3} − 1 are right. The horizontal shift can change how quickly or where along the x‑axis the function starts getting close to the asymptote, but not the asymptote’s y‑value itself. I sometimes think of it as “you arrive at the same floor, just by a different hallway.”
Yup-sliding the input x → x−3 just scoots the graph sideways while the “floor” (the horizontal asymptote) stays put, so only the outside −3 or −1 moves it… though a big sideways shove kind of feels like a tilt in how you approach it. Example: as x→−∞, 2^{x−3}−1 = 2^{-3}·2^{x}−1 → 0−1 = −1 and 2^{x}−3 → 0−3 = −3.
Great question! Think of an exponential graph like a rug lying flat near the floor: sliding it left or right (a horizontal shift) doesn’t change the floor level, but lifting it up or down (a vertical shift) does. For y = 2^x − 3, the base graph 2^x has asymptote y = 0, and subtracting 3 simply drops everything by 3, so the asymptote becomes y = −3. For y = 2^{x−3} − 1, the “x − 3” just delays the same decay, and a quick algebra trick makes this crystal clear: 2^{x−3} = 2^{-3}·2^x = (1/8)·2^x, so y = (1/8)·2^x − 1 still tends to −1 as x → −∞. In limit language: since 2^x → 0 as x → −∞, also 2^{x−3} → 0 because x − 3 → −∞ too; the horizontal shift doesn’t change the zero limit. Simple worked check: at x = −100, 2^{-100} − 3 ≈ −3 and 2^{-103} − 1 ≈ −1, both already hugging their respective asymptotes. General rule: for y = A·b^{x−h} + k with b > 1, the horizontal asymptote is y = k-only the vertical shift k moves it (A and h just change steepness and left/right position).