Base above 1 grows; base below 1 decays.
The two curves are mirror images in the y-axis, because (12)x is the same as 2−x.
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- Graphing Exponential Functions
Graphing Exponential Functions
Every graph of y = b^x passes through (0, 1) and hugs the x-axis on one side.
Choose the point
Which point lies on the graph of y = b^x for every allowed base b?
Anything except zero raised to the power 0 is 1, so b⁰ = 1 for every allowed base and the graph always passes through (0, 1). A is the point every LOGARITHM graph shares, which is the reflection of this one. C works only when b = 1, and that base is not allowed. D can never happen: an exponential output is never zero, which is why the x-axis is an asymptote.
The x-axis is an asymptote, never a crossing.
2⁻¹⁰
≈ 0.000977
tiny, but positive
2⁻³⁰
≈ 0.00000000093
still positive
2x = 0
no solution
never reaches the axis
This is why the range is y > 0 and never y ≥ 0. Halving something forever makes it small, not absent.
Domain is everything; range is everything positive.
domain: all real numbers
any exponent is allowed
range: y > 0
outputs never reach 0
y-intercept: (0, 1)
and no x-intercept at all
"No x-intercept" is a real and complete answer here, not a sign that something went wrong.
Adding outside moves the asymptote with the curve.
y = 2x + 3
everything lifts by 3
asymptote y = 3
not y = 0 any more
y-intercept (0, 4)
1 + 3
range y > 3
the floor moved up
The asymptote moves too
Leaving the asymptote at y = 0 after a vertical shift is the commonest error on this topic, and it makes the range wrong as well. The horizontal line the curve approaches is wherever the constant put it.
Run it backwards: find the base from a point.
If we know a point on the curve, put it into the equation and solve for the missing number.
y = bx passes through (3, 64). Find b.
b³ = 64
x = 3, y = 64
b = 4
the cube root of 64
4³ = 64 ✓
check
y = a · bx passes through (0, 5) and (1, 15). Find a and b.
a · b⁰ = 5, so a = 5
the point at x = 0 gives a at once
5 · b = 15
the point at x = 1
b = 3, so y = 5 · 3x
each step of 1 multiplies by 3
Use the point with x = 0 first: b⁰ = 1, so it hands over a with no work. A base must be positive, so an even power never gives a negative b.
Variations: a decay curve shifted down, a flip that crosses the axis, a sideways shift.
A decay curve shifted down: y = (13)x − 2
base 13 is below 1
decays to the right
asymptote y = −2
the floor moved down 2
y-intercept: 1 − 2 = −1
(0, −1)
range y > −2
above the new floor
Decay changes the direction of the curve, not where the asymptote sits.
A flip that crosses the axis: y = −2x + 8
asymptote y = 8, range y < 8
the minus turns the curve below its asymptote
y-intercept: −1 + 8 = 7
(0, 7)
x-intercept: 2x = 8, so x = 3
set y = 0
After a flip and a lift an exponential graph CAN cross the x-axis, here at (3, 0). Check: −2³ + 8 = 0 ✓.
A sideways shift: y = 2x − 3
asymptote y = 0
a shift inside leaves the floor alone
passes through (3, 1)
the exponent is 0 at x = 3
y-intercept 2−3 = 18
(0, 18)
Only an inside change moves the point (0, 1). Range is still y > 0.
Two points and the asymptote are enough to sketch.
y = 2ˣ
(0, 1) and (1, 2)
asymptote y = 0
y = 2ˣ − 4
(0, −3) and (2, 0)
asymptote y = −4
y = −2ˣ
(0, −1)
flipped below the axis
On the exam
Periodical exams ask for the asymptote, the y-intercept, the range or growth versus decay, often with a sketch; entrance tests add a shift or a flip and ask for the new asymptote or range, or give a point and ask for the base. Find the asymptote first, then put x = 0 for the y-intercept. For a 10-second check, the curve's y-intercept must lie on the correct side of the asymptote — above it unless there is a minus in front.
Now you try
Practice problem
Find the asymptote
y = 3ˣ − 5
3ˣ approaches 0 from above, so 3ˣ − 5 approaches −5 from above. The asymptote is y = −5 and the range is y > −5. A is the asymptote before the shift. B is a vertical line, and exponential graphs never have one. D is the base, which has nothing to do with where the curve flattens out.
Hint: ask what 3ˣ gets close to for very negative x, then apply the −5 to that answer.