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Exponential graphs and e

Sketch exponential graphs, identify asymptotes and transformations, differentiate e to a linear power, and solve coefficient and tangent problems.

Before you startIndex laws, graph transformations, gradients and tangents

01 / What is an exponential?

The variable is in the exponent.

In y = aˣ the base a is a fixed positive number and x can be any real number. The output is always positive. This differs from a power function such as x³, whose base is the variable.

For y = 2ˣ:
x = −2, −1, 0, 1, 2
y = 1/4, 1/2, 1, 2, 4

Increasing x by 1 multiplies the output by a. This constant multiplier, rather than a constant increase, is the key exponential pattern. Negative inputs give reciprocals; they do not make the output negative.

Why use a positive base?

A negative base does not define a real-valued aˣ for every real x. For example, (−2)1/2 is not real. The standard real exponential function uses a > 0.

02 / Growth, decay and the constant case

Read the base before sketching.

a > 1: aˣ increases
0 < a < 1: aˣ decreases
a = 1: aˣ = 1 for every x

Every graph passes through (0,1). For a ≠ 1, its range is y > 0 and y = 0 is a horizontal asymptote: the curve approaches it but never reaches it. The constant graph for a = 1 is the line y = 1 instead.

Choose a base and move the point. For positive x, a larger base gives a larger output; for negative x the order reverses. Bases a and 1/a produce graphs reflected in the y-axis.

The model also shows a tangent gradient. The factor ln(a) is the natural logarithm of a, introduced later in this chapter; when a = e this factor is exactly 1.

Change the base and move the pointMove at your pace
Change the base and move the pointBase a = e. At x = 0, f(x) = 1 and its gradient is 1. The graph is increasing. It passes through (0,1), stays positive and has horizontal asymptote y = 0. The green tangent uses gradient ln(a) times f(x). Displayed decimal values are rounded to four places.f(x) = eˣ-2-10120246810x = 0 f(x) ≈ 1Gradient ≈ 1The graph is increasing.Blue: curve Green: tangent

Base a = e. At x = 0, f(x) = 1 and its gradient is 1. The graph is increasing. It passes through (0,1), stays positive and has horizontal asymptote y = 0. The green tangent uses gradient ln(a) times f(x). Displayed decimal values are rounded to four places.

03 / Transform an exponential graph

Locate the asymptote and intercept first.

For y = A ekx + C:
y-intercept = (0, A + C)
Horizontal asymptote y = C, if A ≠ 0 and k ≠ 0

Multiplying by A scales the heights relative to the asymptote; a negative A also reflects them. Adding C shifts the whole graph vertically. The sign of Ak determines whether this non-constant graph increases or decreases.

y = 3ex/2 − 2:
intercept (0,1), asymptote y = −2, range y > −2

y = 5 − 2eˣ:
intercept (0,3), asymptote y = 5, range y < 5

If k = 0, the graph is the constant y = A + C. If A = 0, it is y = C. Do not apply a non-constant curve description to these degenerate cases.

Inspect a transformed exponentialMove at your pace
Inspect a transformed exponentialFor y = 3eˣ⁄² − 2, A = 3, k = 0.5 and C = -2. The y-intercept is (0,1); the derivative there is 1.5. The graph is increasing. The dashed green line is the horizontal asymptote y = -2. The range is y > -2.y = 3eˣ⁄² − 2-2-1012-10-5051015A = 3, k = 0.5, C = -2Intercept: (0, 1)Horizontal asymptote: y = -2The graph is increasing.

For y = 3eˣ⁄² − 2, A = 3, k = 0.5 and C = -2. The y-intercept is (0,1); the derivative there is 1.5. The graph is increasing. The dashed green line is the horizontal asymptote y = -2. The range is y > -2.

04 / Horizontal shifts and equivalent forms

Keep the complete exponent together.

The graph y = 2x−2 + 3 is y = 2ˣ shifted two units right and three up. Its intercept is (0,13/4), and its horizontal asymptote is y = 3.

e2x+1 = e · e2x
e2(x−3) = e−6e2x

An additive constant inside the exponent becomes a multiplier outside it; it is not a vertical translation. In particular, ex+1 is e times eˣ, while eˣ + 1 is one unit above eˣ.

05 / The special number e

Its exponential has the same height and gradient.

The constant e is approximately 2.71828. Its defining calculus property is:

If f(x) = eˣ, then f′(x) = eˣ

At x = 0, both the height and gradient are 1. At x = 1, both are e. At x = −1, both are 1/e. The graph is increasing everywhere because its derivative is positive.

The optional animation follows a point and its tangent. The derivative is a numerical gradient in the chosen coordinates; a screen angle alone is not a gradient if the axes use different scales.

Watch the height and tangent gradient of eˣ stay equal

Pause, replay or seek freely. The notes explain the same idea and stay in view.

06 / Differentiate exponential expressions

Multiply by the coefficient of x in the exponent.

d(ekx)/dx = k ekx
d(Aekx + C)/dx = Ak ekx

d(4e3x)/dx = 12e3x
d(6e−x/2)/dx = −3e−x/2
d(e2x+1)/dx = 2e2x+1

A constant added outside differentiates to zero. For a sum, differentiate each term. If products can be expanded with index laws, do that first:

eˣ(eˣ + 2) = e2x + 2eˣ
Derivative = 2e2x + 2eˣ

The power rule for xⁿ does not apply to eˣ. Nor may the displayed linear-exponent rule be applied unchanged to ex²; that needs a more general chain rule.

Connecting any positive base to e

Since aˣ = ex ln a, its derivative is (ln a)aˣ. This gives a positive gradient for a > 1, a negative gradient for 0 < a < 1 and zero for a = 1. The logarithm lessons explain the identity used here.

07 / Find a base and coefficient

Divide two point conditions to remove the multiplier.

The curve y = ABˣ, with A > 0 and B > 0, passes through (1,12) and (3,48):

AB = 12, AB³ = 48
B² = 48/12 = 4
B = 2, since B > 0
A = 12/2 = 6

The curve is y = 6 · 2ˣ. If instead (0,80) and (2,20) are given, A = 80 and B² = 1/4, so B = 1/2: a decay graph.

What can an unscaled sketch tell you?

For y = Aekx + C, an asymptote y = 2 and intercept (0,8) give C = 2 and A = 6. If the curve increases, k > 0. Those facts alone do not determine its exact value; another coordinate is needed.

08 / An exponential tangent

Find the point and gradient separately.

For f(x) = ex/3, consider x = 3. The point is (3,e) and the gradient is e/3:

y − e = (e/3)(x − 3)
y = (e/3)x

This tangent passes through the origin even though the exponential curve never does. Its normal has gradient −3/e; use the same point when writing the normal equation.

09 / Combine exponential and polynomial gradients

Perpendicular tangents have gradient product −1.

The curves f(x) = x³ − px + 1 and g(x) = e4x both pass through (0,1). Their tangent gradients there are −p and 4.

(−p)(4) = −1 ⇒ p = 1/4

Use the derivatives, not the function values, in the perpendicularity condition. The two curves meeting at a point does not make their tangents perpendicular automatically.

10 / Your turn

Identify the graph before calculating.

State asymptotes and exact values where possible.

01 · Negative exponent

Find 3⁻² and the y-intercept of y = 3ˣ.

Hint

Use a reciprocal and x = 0.

Worked solution

3⁻² = 1/9; intercept (0,1).

02 · Decay

Describe y = (1/4)ˣ: direction, range and asymptote.

Hint

The base is between 0 and 1.

Worked solution

It decreases, has range y > 0 and asymptote y = 0.

03 · Base one

Does y = 1ˣ approach the x-axis?

Hint

Evaluate it for any real x.

Worked solution

No. It is the constant line y = 1.

04 · A translated graph

Find the y-intercept and asymptote of y = 2x+1 − 5.

Hint

Substitute zero; then inspect the outside shift.

Worked solution

Intercept (0,−3); asymptote y = −5.

05 · A reflected graph

Describe y = 7 − 3e2x.

Hint

Its derivative is negative.

Worked solution

Decreasing; intercept (0,4); horizontal asymptote y = 7; range y < 7.

06 · A derivative

Differentiate 5e−2x + 3eˣ − 4.

Hint

Multiply each exponential by its exponent’s x-coefficient.

Worked solution

−10e−2x + 3eˣ.

07 · Expand first

Differentiate e2x(eˣ + 3).

Hint

Rewrite as e3x + 3e2x.

Worked solution

3e3x + 6e2x.

08 · Two point conditions

y = ABˣ, A,B > 0, passes through (0,5) and (2,45). Find A and B.

Hint

A = 5 and B² = 9.

Worked solution

A = 5, B = 3.

09 · Tangent and normal

Find the tangent and normal to y = e2x at x = 0.

Hint

The point is (0,1), with tangent gradient 2.

Worked solution

Tangent: y = 2x + 1.
Normal: y = 1 − x/2.

10 · Rewrite an exponent

Write e3x−2 as Aekx.

Hint

Split the sum inside the exponent.

Worked solution

A = e⁻², k = 3.

11 · Zero rate

For y = 4ekx + 2, describe the graph when k = 0.

Hint

e⁰ = 1.

Worked solution

The graph is the horizontal line y = 6, with derivative zero.

11 / Recap

Exponential change is multiplicative.

  • Use a positive base; distinguish growth, decay and base 1.
  • Locate the intercept and asymptote before sketching.
  • Keep inside-exponent shifts separate from outside additions.
  • Differentiate Aeᵏˣ by multiplying by k.
  • Use point conditions to recover coefficients.
  • A tangent needs both the curve’s height and its derivative.

Next: exponential growth and decay →

Section 1 of 11 · What is an exponential?