Exponential functions aˣ — graphs, properties, asymptote: Exercises

Before we start, one shared picture. Every graph on this page lives on the same stage:

Figure — Exponential functions aˣ — graphs, properties, asymptote

The red dashed floor is the line $y = 0$ (the $x$ -axis). Every basic $y = a^{x}$ curve hugs it but never lands on it — that floor is the asymptote. Keep this picture in mind for every exercise.

Level 1 — Recognition

Can you name the parts and read off guaranteed facts?

Recall Solution L1.1

What we check: is the variable in the exponent, and is the base a fixed positive number that isn't 1?

(a) $3^{x}$ — ✅ base $3 > 0$ , variable on top. Exponential.
(b) $x^{3}$ — ❌ variable is in the base; this is a cubic power function, not exponential.
(c) $1^{x}$ — ❌ base is $1$ , so $1^{x} = 1$ : a flat line, excluded by $a \neq = 1$ .
(d) $(- 2)^{x}$ — ❌ base negative; $(- 2)^{1/2} = - 2$ is not real. Excluded by $a > 0$ .
(e) $(\frac{1}{5})^{x}$ — ✅ base $\frac{1}{5} > 0$ and $\neq = 1$ . A decay exponential. Answer: (a) and (e).

Recall Solution L1.2

These are the "guaranteed" facts every $a^{x}$ obeys:

(i) $y$ -intercept: put $x = 0$ . Since $a^{0} = 1$ for every base, it's $(0, 1)$ .
(ii) Base-revealing point: put $x = 1$ . Since $a^{1} = a$ , it's $(1, 7)$ — the height at $x = 1$ is the base.
(iii) Range: $y > 0$ (a positive base to any power stays positive).
(iv) Asymptote: $y = 0$ .

Recall Solution L1.3

False. For $x = - N$ (with $N$ a positive whole number) we have $5^{- N} = \frac{1}{5 ^{N}}$ , a positive fraction; as $N$ grows it gets close to $0$ but a nonzero-denominator fraction is never exactly $0$ . The line $y = 0$ is a limit, not a value.

Level 2 — Application

Plug in, compute, and plot.

Recall Solution L2.1

Use $2^{- N} = 1/ 2^{N}$ for negatives, direct multiplication for positives. $2^{- 3} = \frac{1}{8}, 2^{- 1} = \frac{1}{2}, 2^{0} = 1, 2^{1} = 2, 2^{3} = 8.$ As $x \to - \infty$ , values $\frac{1}{8}, \dots$ shrink toward $0^{+}$ → asymptote $y = 0$ .

Recall Solution L2.2

What we do: a negative exponent means "reciprocal," so flip the base: $(\frac{1}{3})^{- 2} = (\frac{3}{1})^{2} = 3^{2} = 9.$ Why: $a^{- n} = 1/ a^{n}$ from Index laws. Reciprocal of $\frac{1}{3}$ is $3$ ; then square. Answer: $9$ .

Recall Solution L2.3

This is a vertical shift up by 3 (a graph transformation).

$y$ -intercept: $x = 0 \Rightarrow 2^{0} + 3 = 1 + 3 = 4$ , so $(0, 4)$ .
Asymptote: the base curve's floor $y = 0$ rises with everything else to $y = 3$ .
Range: since $2^{x} > 0$ , we get $2^{x} + 3 > 3$ , so range $y > 3$ .

Level 3 — Analysis

Explain the WHY behind the behaviour.

Recall Solution L3.1

Shared point: $a^{0} = 1$ for every base, so both hit $(0, 1)$ . Steepness: the height at $x = 1$ is the base — $2^{x}$ reaches $2$ , $5^{x}$ reaches $5$ . Rising to a bigger height over the same run means a steeper climb, and on the left $5^{x}$ hugs the asymptote more tightly. Meeting point: we want the $x$ where $2^{x} = 5^{x}$ . Divide both sides by $2^{x}$ ; this is legal because $2^{x} > 0$ (never zero), so we are dividing by a nonzero quantity. That gives $\frac{5 ^{x}}{2 ^{x}} = 1$ , and since $\frac{5 ^{x}}{2 ^{x}} = (\frac{5}{2})^{x}$ (index law $\frac{a ^{x}}{b ^{x}} = (a / b)^{x}$ ), we get $(\frac{5}{2})^{x} = 1$ . A base bigger than 1 raised to a power equals 1 only at exponent $0$ , so $x = 0$ . They cross only at $(0, 1)$ .

The figure below shows this: the two curves funnel through the single orange dot at $(0, 1)$ , and to the right of it $5^{x}$ (orange) climbs far faster than $2^{x}$ (blue); to the left it hugs the red dashed floor more tightly.

Figure: $y = 2^{x}$ (blue) and $y = 5^{x}$ (orange) meet only at the green dot $(0, 1)$ . The steeper orange curve reaches height $5$ at $x = 1$ while blue reaches only $2$ ; both flatten toward the red dashed asymptote $y = 0$ on the left.

Recall Solution L3.2

Rewrite: $(\frac{1}{2})^{x} = (2^{- 1})^{x} = 2^{- x}$ (index law $(a^{m})^{n} = a^{mn}$ ). Replacing $x$ by $- x$ inside a function reflects its graph in the $y$ -axis. So $(\frac{1}{2})^{x}$ is $2^{x}$ read backwards: growth becomes decay. Asymptote: as $x \to + \infty$ , $2^{- x} = 1/ 2^{x} \to 0^{+}$ . Same floor $y = 0$ , approached from the other side.

The figure below makes the mirror explicit: fold the picture along the dotted $y$ -axis and the blue growth curve lands exactly on the orange decay curve.

Figure: $y = 2^{x}$ (blue, growth) and $y = (1/2)^{x} = 2^{- x}$ (orange, decay) are mirror images across the dotted gray $y$ -axis. Both pass through $(0, 1)$ and flatten toward the red dashed asymptote $y = 0$ — blue on the left, orange on the right.

Recall Solution L3.3

Let $N$ be a positive whole number ( $N \in {1, 2, 3, \dots}$ ) and write $x = - N$ ; letting $x \to - \infty$ is the same as letting $N \to + \infty$ . Then $2^{- N} = \frac{1}{2 ^{N}}$ . Since $2 > 1$ , repeated multiplication makes $2^{N} \to \infty$ as $N \to \infty$ , so $\frac{1}{2 ^{N}} \to 0^{+}$ : arbitrarily close to 0. But for any finite $N$ , $2^{N}$ is a finite positive number, so $1/ 2^{N}$ is a positive fraction — never $0$ . Hence $y = 0$ is approached but never reached: a true asymptote.

Level 4 — Synthesis

Combine transformations, laws, and reasoning.

Recall Solution L4.1

Build it in two moves:

Reflect in $x$ -axis: $y = 3^{x} \to y = - 3^{x}$ . Every output flips sign; the range $y > 0$ becomes $y < 0$ ; asymptote still $y = 0$ ; curve now points downward.
Shift up 5: $y = - 3^{x} + 5$ . Add 5 to everything.

$y$ -intercept: $x = 0 \Rightarrow - 3^{0} + 5 = - 1 + 5 = 4$ , so $(0, 4)$ .
Asymptote: the floor $y = 0$ becomes a ceiling $y = 5$ (curve now sits below it).
Range: $- 3^{x} < 0$ , so $- 3^{x} + 5 < 5$ , giving range $y < 5$ .

Recall Solution L4.2

First: write $\frac{1}{8}$ as a power of the same base. $\frac{1}{8} = \frac{1}{2 ^{3}} = 2^{- 3}$ . So $2^{x} = 2^{- 3} \Rightarrow x = - 3$ . Second: put both sides over base $3$ . $9 = 3^{2}$ and $27 = 3^{3}$ , so $(3^{2})^{x} = 3^{3} \Rightarrow 3^{2 x} = 3^{3} \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2}$ . (This "match the bases" trick is exactly what logarithms automate later.) Answers: $x = - 3$ and $x = \frac{3}{2}$ .

Recall Solution L4.3

"Triples every hour" = multiply by $3$ each step, so the model is $N (t) = 50 \cdot 3^{t} .$

After $3$ hours: $N (3) = 50 \cdot 3^{3} = 50 \cdot 27 = 1350$ cells.
One hour before start, $t = - 1$ : $N (- 1) = 50 \cdot 3^{- 1} = \frac{50}{3} \approx 16.67$ cells. The negative time uses $3^{- 1} = 1/3$ — the same reciprocal rule that gives decay, showing growth "run backwards."

Level 5 — Mastery

Prove and construct from first principles.

Recall Solution L5.1

Here we use the index law $a^{m + n} = a^{m} a^{n}$ for integer exponents $m, n$ — the case already established in Index laws from repeated multiplication. The whole point is to extend it downward to $0$ and to negatives by demanding it keep holding. $a^{0} = 1$ : set $m = 0, n$ a positive integer: $a^{0 + n} = a^{0} \cdot a^{n}$ , i.e. $a^{n} = a^{0} \cdot a^{n}$ . Since $a^{n} > 0$ we may divide by it: $1 = a^{0}$ . ✔ $a^{- n} = 1/ a^{n}$ : set $m = - n, n$ a positive integer: $a^{- n + n} = a^{- n} \cdot a^{n}$ , i.e. $a^{0} = a^{- n} \cdot a^{n}$ . But $a^{0} = 1$ , so $a^{- n} \cdot a^{n} = 1 \Rightarrow a^{- n} = \frac{1}{a ^{n}}$ . ✔ Both facts are forced by demanding one law survive to all integer exponents — nothing is invented.

Recall Solution L5.2

What we need: $a^{3} = 64$ . Take the cube root (the inverse of cubing): $a = 364 = 4$ , and $4 > 1$ ✔. So the function is $y = 4^{x}$ . At $x = - 1$ : $4^{- 1} = \frac{1}{4}$ . Answers: $a = 4$ , and $y (- 1) = \frac{1}{4}$ .

Recall Solution L5.3

Set up the condition at $x = 2$ : $a^{2} = 9 b^{2}$ . With $b = 2$ : $a^{2} = 9 \cdot 4 = 36$ . Take the positive root (a growth base is $> 1$ ): $a = 36 = 6$ . Check: $a^{2} / b^{2} = 36/4 = 9$ ✔, and $6 > 1$ so it is genuinely growth. Answer: $a = 6$ .

Active recall

Recall Quick self-test (try to answer, then check)

Why do $2^{x}$ and $5^{x}$ meet only at $(0, 1)$ ?
How does reflecting $3^{x}$ in the $x$ -axis change its range and asymptote?
Solve $9^{x} = 27$ by matching bases.
From the index law alone, prove $a^{0} = 1$ .

Recall Answers

Dividing $2^{x} = 5^{x}$ by $2^{x}$ (which is never $0$ ) gives $(5/2)^{x} = 1$ , true only at $x = 0$ .
Reflecting flips the range to $y < 0$ ; asymptote stays $y = 0$ (until you also shift).
$9 = 3^{2}, 27 = 3^{3}$ , so $3^{2 x} = 3^{3} \Rightarrow x = \frac{3}{2}$ .
Set $m = 0$ in $a^{m + n} = a^{m} a^{n}$ : $a^{n} = a^{0} a^{n}$ , divide by $a^{n} > 0$ to get $a^{0} = 1$ .

Connections

Exponential functions aˣ — graphs, properties, asymptote — the parent this page drills.
Index laws — every reciprocal/base-matching move above.
Graph transformations — the shift/reflect problems (L2.3, L4.1).
Exponential growth and decay models — the bacteria model (L4.3).
Logarithms as the inverse of exponentials — automates the base-matching of L4.2.
The number e and natural exponential eˣ — the special growth base.