Every smartphone in your pocket contains billions of transistors — and that number has doubled roughly every two years for over five decades. This is exponential growth in action. In this lesson, you will learn to sketch, transform, and interpret the graphs of exponential functions, connecting the abstract mathematics to the hardware that powers modern technology.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, an extend task and success-criteria proof.
Prediction — Sketch $y = 2^x$ and $y = (\frac{1}{2})^x$ without calculating points. What do you notice about their symmetry?
Make a prediction before reading on. The relationship between these two graphs reveals a fundamental property of exponential functions.
Type your initial prediction below — you will revisit this at the end of the lesson.
Write your initial prediction in your book. You will revisit it at the end of the lesson.
Core Content
The basic exponential function $y = a^x$ (where $a > 0$, $a \neq 1$) has a characteristic shape that depends on whether the base is greater than 1 or between 0 and 1.
Both families share the horizontal asymptote $y = 0$ and the $y$-intercept $(0, 1)$. The only difference is the direction of growth.
For growth functions ($a > 1$), larger bases produce steeper graphs. Consider three functions on the same axes:
$y = 2^x$, $y = 3^x$, $y = 10^x$
At $x = 1$: the outputs are $2$, $3$, and $10$ respectively. At $x = 2$: they are $4$, $9$, and $100$. The gap widens dramatically because exponential growth is multiplicative, not additive.
For decay functions ($0 < a < 1$), the opposite holds: smaller bases decay faster. The function $y = (\frac{1}{10})^x$ drops toward zero much more rapidly than $y = (\frac{1}{2})^x$.
Show three exponential growth curves on one set of axes with $x$ from -2 to 3 and $y$ from 0 to 12. Curve A (blue): $y = 2^x$ passing through (-1, 0.5), (0, 1), (1, 2), (2, 4). Curve B (green): $y = 3^x$ passing through (-1, 0.33), (0, 1), (1, 3), (2, 9). Curve C (orange): $y = 10^x$ passing through (-1, 0.1), (0, 1), (1, 10). Label each curve. Mark the common $y$-intercept $(0, 1)$ with a dot and the horizontal asymptote $y = 0$ as a dashed line. Show that C is steepest, then B, then A.
The general transformed exponential function is:
$$y = A \cdot a^{x-h} + k$$
Each parameter changes the graph in a predictable way:
Sketch $y = 2^{x-1} + 3$, showing key features.
The $y$-intercept, the horizontal asymptote, and two other points on the curve.
Start with y = 2^x.
Shift right 1: y = 2^(x-1)
Shift up 3: y = 2^(x-1) + 3
Asymptote: y = 3 (shifted up 3)
When x = 1: y = 2^0 + 3 = 1 + 3 = 4
When x = 2: y = 2^1 + 3 = 2 + 3 = 5
When x = 0: y = 2^(-1) + 3 = 0.5 + 3 = 3.5
y-intercept: (0, 3.5)
Horizontal asymptote: $y = 3$
The curve passes through $(0, 3.5)$, $(1, 4)$, and $(2, 5)$.
It rises from the asymptote $y = 3$ on the left, passing through these points, and increases to the right.
Sketch $y = 3^x - 2$, labelling the asymptote and the $y$-intercept.
Answer:
Asymptote: $y = -2$ (shifted down 2).
$y$-intercept: when $x = 0$, $y = 3^0 - 2 = 1 - 2 = -1$, so $(0, -1)$.
Another point: when $x = 1$, $y = 3 - 2 = 1$, so $(1, 1)$.
When $x = -1$, $y = 3^{-1} - 2 = \frac{1}{3} - 2 = -\frac{5}{3}$, so $(-1, -\frac{5}{3})$.
$y = a^x$
$y$-intercept: $(0, 1)$
Asymptote: $y = 0$
$y = a^{-x} = (\frac{1}{a})^x$
Reflection of $y = a^x$ in the $y$-axis
$y = -a^x$
Reflection in the $x$-axis
Range: $y < 0$
$y = a^{x-h} + k$
Shift right $h$, up $k$
Asymptote: $y = k$
For each function, find the horizontal asymptote, the $y$-intercept, and one other point. Then describe the transformation from $y = 2^x$.
A smartphone manufacturer observes that transistor counts follow an exponential pattern, doubling every 24 months.
Look back at what you wrote in the Think First section. The graphs of $y = 2^x$ and $y = (\frac{1}{2})^x = 2^{-x}$ are reflections of each other in the $y$-axis. Both pass through $(0, 1)$, but as $x \to +\infty$, $y = 2^x$ shoots upward while $y = (\frac{1}{2})^x$ decays toward zero.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. For $f(x) = 2^{x+1} - 3$, find (a) the horizontal asymptote, (b) the $y$-intercept, (c) the $x$-intercept. 3 MARKS
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Answer in your workbook.
9. The graph of $y = a^x$ passes through $(2, 16)$. Find $a$, then sketch the graph showing key features. 3 MARKS
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Answer in your workbook.
10. Explain why the transformation $y = 2^{x-3}$ represents a horizontal shift right by 3 units, even though the number inside the bracket is subtracted. Use specific points to justify your answer. 3 MARKS
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Answer in your workbook.
1. $y = 2^{x-2}$: Asymptote $y = 0$; $y$-intercept $(0, 2^{-2}) = (0, \frac{1}{4})$; point $(2, 1)$. Shifted right 2 units.
2. $y = 2^x + 4$: Asymptote $y = 4$; $y$-intercept $(0, 5)$; point $(1, 6)$. Shifted up 4 units.
3. $y = 2^{x+1} - 3$: Asymptote $y = -3$; $y$-intercept $(0, 2^1 - 3) = (0, -1)$; point $(-1, -2)$. Shifted left 1, down 3.
4. $y = -2^x$: Asymptote $y = 0$; $y$-intercept $(0, -1)$; point $(1, -2)$. Reflected in the $x$-axis.
1. $T(t) = 1 \cdot 2^{t/2}$ billion transistors, since doubling every 2 years means base $2^{1/2}$ per year, so $t$ years gives $(2^{1/2})^t = 2^{t/2}$.
2. $T(0) = 1$ billion. $T(2) = 2$ billion. $T(4) = 4$ billion. Graph rises from $(0, 1)$ through $(2, 2)$ and $(4, 4)$, asymptotic to $y = 0$ as $t \to -\infty$.
3. The graph never intersects the $t$-axis because $2^{t/2} > 0$ for all real $t$. Physically, transistor counts are always positive; even the earliest chips had transistors, and the model only makes sense for $T > 0$.
Q8 (3 marks): (a) Horizontal asymptote: $y = -3$ [1]. (b) $f(0) = 2^{0+1} - 3 = 2 - 3 = -1$, so $y$-intercept is $(0, -1)$ [1]. (c) $2^{x+1} - 3 = 0 \Rightarrow 2^{x+1} = 3 \Rightarrow x+1 = \log_2 3 \Rightarrow x = \log_2 3 - 1 \approx 1.585 - 1 = 0.585$ [1].
Q9 (3 marks): $a^2 = 16 \Rightarrow a = 4$ (base must be positive) [1]. So $y = 4^x$. Key features: $y$-intercept $(0, 1)$; asymptote $y = 0$; point $(2, 16)$; point $(1, 4)$ [1]. Sketch: rising curve from left asymptote through $(0, 1)$ and $(2, 16)$, getting steeper [1].
Q10 (3 marks): In $y = 2^x$, the key point is $(0, 1)$ because $2^0 = 1$ [1]. In $y = 2^{x-3}$, the exponent is zero when $x - 3 = 0$, i.e. $x = 3$. So $2^{3-3} = 2^0 = 1$, meaning the point $(0, 1)$ has moved to $(3, 1)$ [1]. Similarly, $(1, 2)$ moves to $(4, 2)$. Every point shifts 3 units to the right, confirming a horizontal shift right by 3 [1].
Climb platforms using your knowledge of exponential graphs, transformations, asymptotes, and reflections. Pool: lesson 2.
Tick when you've finished all activities and checked your answers.