I’m revising exponential graphs to strengthen my fundamentals, and my brain keeps doing that thing where it tries to fold the graph like origami and then I can’t tell which corner goes where. I’m especially stuck on how to think about transformations that happen inside the exponent.
Example 1: Compare f(x) = 2^x with g(x) = 2^{3x}.
– I learned that multiplying x by 3 should be a horizontal compression by a factor of 1/3. But g(x) = 2^{3x} is also equal to 8^x, which to me feels like “same shape, just a different base.” So… which mental model should I use when sketching? Is it “compressed horizontally” or “steeper because the base is bigger,” or are those literally the same idea in disguise?
– Also, I think the y-intercept is still 1, because g(0) = 2^{0} = 1, even after the 3x inside. That seems right, but I keep second-guessing myself because rewriting it as 8^x makes me feel like something should shift.
Example 2: h(x) = -3*(1/2)^{x – 4} + 5.
– My read: base is 1/2 so it’s a decay shape, the -3 flips it over the x-axis and stretches vertically, shift right by 4, up by 5.
– Horizontal asymptote should be y = 5 (I think?).
– Domain: all real x. Range: (-∞, 5). For the y-intercept, h(0) = -3*(1/2)^{-4} + 5 = -3*16 + 5 = -43. That all feels correct, but the “decay + reflection” combo is messing with my intuition about which end of the graph snuggles up to the asymptote.
Could someone explain clearly how to think about the “multiply x inside the exponent” idea versus “just change the base,” and when one viewpoint is better for quick sketching? And can you sanity-check my asymptote/range/intercept for h(x) and help me lock in the left/right end behavior without mental gymnastics?
Any help appreciated!
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3 Responses
Great question! 2^{3x} is a horizontal squeeze of 2^x by 1/3, and that’s exactly the same as rewriting it as 8^x (like playing a song at 3× speed vs turning up a “steeper” volume knob-either way it rises faster), so the y-intercept stays 1 and the asymptote is still y=0; for h(x) = -3(1/2)^{x-4}+5, your read is right: asymptote y=5, domain all reals, range (-∞, 5), y-intercept -43, with h(x) → 5 from below as x → ∞ and h(x) → -∞ as x → -∞. For a quick refresher, see: https://www.khanacademy.org/math/algebra/exponential-growth-decay.
For 2^{3x}, both descriptions are the same: 2^{3x} = (2^3)^x = 8^x, and multiplying x by 3 is a horizontal compression by factor 1/3. The “bigger base → steeper” view and the “horizontal squeeze” view describe the same graph. For quick sketching, use the squeeze when you want to move known points: for 2^x the point (1,2) becomes (1/3,2) because 2^{3(1/3)}=2, so doubling happens every 1/3 instead of every 1 unit. Use the base-change view when comparing growth rates. The horizontal asymptote stays y=0 and g(0)=2^0=1, so the y-intercept is still 1 (rewriting as 8^x doesn’t shift anything). For h(x) = -3(1/2)^{x-4} + 5, your read is right: decay base, vertical reflection and stretch, right 4, up 5. The horizontal asymptote is y=5; domain is all real x; range is (-∞, 5). The y-intercept h(0) = -3·(1/2)^{-4}+5 = -43 is correct. For end behavior: as x→∞, (1/2)^{x-4}→0, so h(x)→5 from below; as x→-∞, (1/2)^{x-4}→∞, so h(x)→-∞. The graph is increasing overall (a reflected decay). Hope this helps!
For 2^x vs 2^{3x}, those two mental pictures are actually the same idea wearing different hats. Multiplying x by 3 is a horizontal squeeze by 1/3 because every “time” it takes f(x)=2^x to reach a certain height now happens three times sooner in g(x)=2^{3x}. But algebraically 2^{3x}=(2^3)^x=8^x, which just says “same exponential shape, faster growth per 1-step in x.” I like the time-lapse analogy: you can either speed up the clock (compress the x-axis) or use a faster-growing plant; the video you watch looks identical. For quick sketching, use the squeeze view when you’re mapping known points of 2^x (e.g., the point (1,2) on 2^x corresponds to (1/3,2) on 2^{3x}), and use the “change the base” view when you want instant feel for growth rate. In any case, the y-intercept stays 1 (g(0)=1), the horizontal asymptote stays y=0, and the range stays (0,∞). When I first learned this, I kept flipping between “steeper” and “squeezed” until a teacher said, “Pick whichever makes the sketch faster today,” and that finally clicked for me.
For h(x)=-3·(1/2)^{x-4}+5, your read is spot on: start with decay (1/2)^x, shift right 4, reflect over the x-axis and stretch by 3, then shift up 5. The horizontal asymptote is y=5. As x→+∞, (1/2)^{x-4}→0, so h(x)→5 from below; as x→−∞, (1/2)^{x-4}→+∞, so the reflected-and-stretched part dives to −∞, and adding 5 still gives −∞. So the graph “snuggles” up to y=5 on the right-hand side, and drops without bound on the left. Domain is all real numbers, range is (−∞,5). Your y-intercept check h(0)=−3·(1/2)^{−4}+5=−48+5=−43 is correct. I used to picture this like flipping a downhill ramp upside down and then lifting the whole ramp up by 5 units-the end far to the right just levels off near y=5, while the left side plunges downward.