Exponential functions aˣ — graphs, properties, asymptote: Question bank

True or false — justify

Every graph

y = a^{x}

passes through

(0, 1)

, whatever the base.

True —

a^{0} = 1

holds for every allowed base

a

, so the height at

x = 0

is always exactly

1

; the base only changes steepness, not this shared anchor.

y = a^{x}

can output a negative number if

x

is negative.

False — a negative exponent means

a^{- N} = 1/ a^{N}

, a positive number divided by a positive number; the exponent being negative flips it toward small, not toward negative.

The graph of

y = a^{x}

eventually touches the

x

-axis on the far left (for

a > 1

False — it gets arbitrarily close to

y = 0

but never reaches it, because

1/ a^{N}

is a positive fraction that is never zero;

y = 0

is a limit, not a value.

y = 5^{x}

starts higher up the

y

-axis than

y = 2^{x}

because

5 > 2

False — both cross the

y

-axis at

(0, 1)

since

5^{0} = 2^{0} = 1

; the larger base only makes

5^{x}

climb steeper to the right and hug the asymptote tighter on the left.

y = (1/2)^{x}

is an increasing function because you can write it with a positive base

1/2

False —

0 < 1/2 < 1

means repeated multiplication by a fraction shrinks the output, so it decreases; equivalently

(1/2)^{x} = 2^{- x}

2^{x}

reflected in the

y

-axis.

Both

y = 2^{x}

and

y = (1/2)^{x}

have the same horizontal asymptote.

True — both approach

y = 0

(one as

x \to - \infty

, the other as

x \to + \infty

); reflecting in the

y

-axis keeps the floor

y = 0

fixed.

y = x^{2}

and

y = 2^{x}

are both exponential functions.

False —

x^{2}

has the variable in the base (a parabola/power function);

2^{x}

has the variable in the exponent, and only that is exponential.

The range of

y = 2^{x} - 1

y > 0

False — shifting the whole curve down by

1

drags the floor to

y = - 1

, so the range is

y > - 1

(the asymptote moved with the curve).

Spot the error

a^{x} = 0

when

x

is very large and negative, so the graph meets

y = 0

." Where's the flaw?

The step "very close to

0

" was silently upgraded to "equal to

0

a^{x} = 1/ a^{N}

shrinks without limit but a fraction with a finite denominator is never

0

; approaching is not reaching.

"Since

1^{x}

is positive and never zero,

1^{x}

is a valid exponential function." What's wrong?

1^{x} = 1

for all

x

is a flat horizontal line, not a genuine exponential — that is exactly why the definition forbids

a = 1

a^{x}

with

a = - 2

gives

(- 2)^{x}

, a perfectly good exponential." Why is this rejected?

For fractional exponents like

x = 1/2

we'd need

- 2

, which is not real, so the function would have holes everywhere; hence the rule

a > 0

A student writes "for

y = 2^{x} + 3

the asymptote is still

y = 0

". Correct them.

Adding

3

raises every output including the floor, so the asymptote shifts up to

y = 3

and the range becomes

y > 3

(see Graph transformations).

2^{x}

grows faster than

3^{x}

because it doubles and doubling is dramatic." Diagnose.

Bigger base means faster growth: at

x = 1

3^{x} = 3 > 2 = 2^{x}

, and the gap widens; "doubling" sounds dramatic but

3^{x}

triples each step and overtakes for all

x > 0

"To sketch

y = a^{x}

I only need the point

(0, 1)

." What's missing?

One point can't reveal direction or steepness; you need a second anchor

(1, a)

(which is the base) plus a negative

x

value to see the curve flattening toward the asymptote.

(1/3)^{x}

and

3^{x}

are unrelated curves." Correct this.

They are mirror images:

(1/3)^{x} = 3^{- x}

, so one is the reflection of the other in the

y

-axis — decay is growth run backwards.

Why questions

Why must the base be strictly positive rather than just non-zero?

A zero base gives

0^{x}

(undefined for

x \leq 0

) and a negative base gives non-real values at fractional exponents; only

a > 0

keeps

a^{x}

real for every real

x

Why does every exponential pass through

(0, 1)

regardless of base?

The index law forces

a^{0} = 1

: from

a^{0} \cdot a^{n} = a^{0 + n} = a^{n}

we divide to get

a^{0} = 1

, and this is base-independent (see Index laws).

Why can we even talk about

a^{2}

when

2

isn't a fraction?

We squeeze

2

between rational approximations

1.41, 1.414, \dots

and take the limit of

a^{1.41}, a^{1.414}, \dots

; continuity fills the gaps between the rational powers.

Why is the range

y > 0

and not

y \geq 0

a^{x}

is always a positive number and the value

0

is only ever approached as a limit, never attained, so

0

itself is excluded from the range.

Why does a larger base give a steeper curve rather than a higher starting point?

All curves are pinned at

(0, 1)

; the base equals the height at

x = 1

, so a larger base means a taller value at

x = 1

and therefore a steeper rise — steepness, not intercept, is what the base controls.

Why does reflecting

y = a^{x}

in the line

y = x

(not the

y

-axis) give something new?

Reflecting in

y = x

swaps input and output, producing the inverse — the logarithm

lo g_{a} x

— whereas reflecting in the

y

-axis only turns growth into decay (see Logarithms as the inverse of exponentials).

Why is

e^{x}

singled out among all the

a^{x}

e

is the unique base for which the curve's slope equals its own height at every point, which makes calculus with it clean (see The number e and natural exponential eˣ).

Edge cases

What does

y = a^{x}

look like at exactly

a = 1

, and why is it excluded?

It collapses to the constant line

y = 1

; excluded because it has no growth or decay — nothing exponential about a flat line.

a

shrinks from just above

1

down toward

1

, what happens to the growth curve?

It flattens toward the horizontal line

y = 1

: growth becomes gentler and gentler until, in the limit

a \to 1

, there is no growth at all.

For

y = a^{x}

with

0 < a < 1

, on which side does the curve blow up, and on which side does it hug

y = 0

It blows up as

x \to - \infty

and hugs

y = 0

x \to + \infty

— the exact mirror of the

a > 1

case, since

(1/ a)^{x} = a^{- x}

What is

a^{0}

when

a

is tiny, say

a = 0.001

, and does the

(0, 1)

rule still hold?

Yes —

0.00 1^{0} = 1

, because

a^{0} = 1

for any positive base; the intercept is immovable no matter how small (or large) the base is.

For the decay curve

y = (1/2)^{x}

, is there any finite

x

where the output equals zero?

No —

(1/2)^{x} = 2^{- x}

stays strictly positive for every real

x

; it only tends to

0

x \to + \infty

, never landing on it.

If you shift

y = 3^{x}

down by exactly its own asymptote value, where's the new floor?

Its asymptote is

y = 0

, so shifting down by

0

changes nothing; but shifting down by any positive

c

moves the floor to

y = - c

and lets the curve genuinely cross the

x

-axis once.

Active recall

Recall Rapid-fire self-test

Can $a^{x}$ ever be zero or negative? Justify in one sentence.
Which single point do all exponential graphs share, and why is it base-independent?
What goes wrong if $a < 0$ ? If $a = 1$ ?
How does the asymptote move under $y = 2^{x} + 3$ versus $y = 2^{x} - 1$ ?
Why is $(1/ a)^{x}$ the mirror of $a^{x}$ ?

Connections

Exponential functions aˣ — graphs, properties, asymptote — the parent this bank drills.
Index laws — the source of $a^{0} = 1$ and $a^{- n} = 1/ a^{n}$ behind half these traps.
Graph transformations — why shifts move the asymptote and range.
Logarithms as the inverse of exponentials — the $y = x$ reflection edge case.
The number e and natural exponential eˣ — the special base singled out.
Exponential growth and decay models — where growth-vs-decay confusions bite in practice.