| letter | what it does to the picture | moves the asymptote? |
|---|---|---|
| b | sets how fast — climbing if b > 1, falling if 0 < b < 1 | no |
| a | stretches the curve away from the asymptote by a factor of |a|, and flips it to the other side if a < 0 | no |
| h | slides the whole picture sideways — right when h > 0, left when h < 0 | no |
| k | lifts the whole picture straight up by k | yes — to y = k |
An asymptote is a claim about a gap: the distance between the curve and a line goes to zero. So write the gap down. Subtract k from both sides:
Now read the right-hand side. A power of a positive base is never zero and never negative, so the gap never actually closes and never changes sign. But as the exponent x − h marches off toward −∞ (or toward +∞ when 0 < b < 1), that power shrinks past every number you can name. The gap heads to 0 without ever arriving. That is precisely what a horizontal asymptote is, and the line it belongs to is y = k.
Notice what the argument needed: only that a is some fixed number and h is some fixed number. Their values never entered. That is why they cannot touch the asymptote — the proof never asked what they were.
Take y = 2x and stretch it by a factor of 50. Fifty is a lot. It is not enough:
| x | 2x | 50·2x |
|---|---|---|
| −5 | 1/32 = 0.03125 | 50/32 = 1.5625 |
| −10 | 1/1024 ≈ 0.000977 | 50/1024 ≈ 0.0488 |
| −20 | ≈ 0.00000095 | ≈ 0.0000477 |
| −30 | ≈ 0.00000000093 | ≈ 0.0000000466 |
The stretched column is always 50 times bigger and it is still collapsing to zero. Fifty times a vanishing quantity is a vanishing quantity. The stretch buys the curve a head start, not a landing spot. Switch the figure above to The gap and drag a — you will see every gap bar rescale together while the line they are measured from does not budge.
The template subtracts h, so a plus sign in the exponent means a negative h:
Do not memorize left and right. Ask instead: which x makes the exponent zero? That is the only x where b0 = 1, and it is where the parent curve's starting point ended up. For 2x−3 the exponent is zero at x = 3, so the anchor is at x = 3 — three units right. For 2x+1 it is zero at x = −1 — one unit left. The question answers itself and there is nothing to forget.
Split the exponent apart. This is not a special rule — it is just counting factors of b:
Read that again, because it is strange. The horizontal shift turned into a constant multiplier. Concretely:
So for an exponential, "shift right 3" and "compress vertically to one eighth" describe the same curve. You cannot tell them apart by looking, because there is nothing to tell apart. And that instantly explains the asymptote result: a horizontal shift is a vertical stretch in disguise, and we already proved a stretch cannot move the asymptote.
This is genuinely a property of exponentials and not of functions in general. Try it on a parabola and it falls apart at once: (x − 3)2 = x2 − 6x + 9, which is not any constant times x2. Shifting a parabola sideways genuinely moves it sideways. Shifting an exponential sideways just rescales it.
Stretch first and then lift, or lift first and then stretch? They are not the same, and the difference lands squarely on the asymptote:
The second one lifted the asymptote to 2 and then stretched that away from the x-axis as well, tripling it to 6. When a function is handed to you in the form a·bx−h + k, the k sits outside everything and is added last — so the reading is always stretch, then lift, and the asymptote is plain k with nothing done to it.
The gap equation settles this too. Flipping the sign of a flips the sign of f(x) − k, which is the signed distance from the line y = k. Same size, other side:
Because the gap can never be zero and never changes sign for a fixed a, the curve is trapped on one side of y = k forever. It cannot cross the line, cannot touch it, cannot loop back. That single fact hands you the range without any further work.
Domain: every real number. There is nothing you are forbidden to put in an exponent — no division by zero to dodge, no negative under a square root. So the domain is (−∞, ∞) with no excluded values, every time.
Range: one-sided, decided entirely by the sign of a. If a > 0 the range is y > k; if a < 0 it is y < k. The value y = k itself is never attained — use a strict inequality, and an open circle if you are sketching the boundary.
The y-intercept always exists. Put x = 0 in: f(0) = a·b−h + k = a/bh + k. Since 0 is in the domain, there is nothing to check.
The x-intercept usually does not. Setting the output to zero gives a·bx−h = −k, so bx−h = −k/a. A power of b is strictly positive, so a solution exists only when −k/a > 0 — that is, only when a and k have opposite signs. Same signs, or k = 0, and there is no x-intercept at all; do not go looking for one. The picture agrees: if a and k are both positive, the curve is stuck above a line that is itself above the x-axis, so it can never reach down to it.
When a solution does exist and −k/a happens to be a recognizable power of b, you can finish by matching exponents — that is what happens in two of the four worked functions. When it is not a nice power, the exponent has to be dug out with logarithms, which is the next chapter's job.
Drag a to exactly 0 and the graph goes flat. That is honest arithmetic, not a glitch: y = 0·bx−h + k = k, a horizontal line. There is no curve, no asymptote to approach — the graph is the line. This is why the definition insists on a ≠ 0, alongside b > 0 and b ≠ 1. Each restriction is there to rule out a case where the shape collapses.