Exponential Functions – OCR FSMQ Additional Maths

1 The Function $ka^x$

EL1: know and use the function $ka^x$ and its graph, where $a$ is positive. The defining feature is that the variable is in the index.

The two parameters

$y = k\,a^x$
$k$ is the value when $x = 0$ — the starting amount
$a$ is the multiplier per unit increase in $x$

What makes it exponential. Each time $x$ goes up by $1$, $y$ is multiplied by $a$. Compare a linear function, where each step adds a fixed amount. Multiplying repeatedly is what produces the explosive behaviour.

$x$	$y = 3x$ (linear)	$y = 3^x$ (exponential)
$0$	$0$	$1$
$1$	$3$	$3$
$2$	$6$	$9$
$3$	$9$	$27$
$5$	$15$	$243$
$10$	$30$	$59\,049$

The linear one wins at first and then loses forever. At $x = 1$ they are equal; by $x = 10$ the exponential is nearly two thousand times larger. That crossover is the whole reason exponential models are taken seriously.

2 The Graph

Features of $y = ka^x$ for $k > 0$

Passes through $(0, k)$ · never crosses the $x$-axis
$a > 1$: growth, rising ever faster · $0 < a < 1$: decay, falling ever more slowly
The $x$-axis is a horizontal asymptote

There is never an $x$-intercept. $ka^x$ is positive for every real $x$ when $k$ and $a$ are positive. A sketch showing the curve meeting or crossing the $x$-axis is wrong.

Worked Example 1 — Sketching a growth curve

Sketch $y = 4 \times 2^x$, marking the intercept and describing both ends.

①$y$-intercept: $x = 0$ gives $y = 4 \times 1 = 4$, so $(0, 4)$.

②A few more points: $x=1 \to 8$; $x=2 \to 16$; $x=3 \to 32$.

③Negative side: $x=-1 \to 2$; $x=-2 \to 1$; $x=-3 \to 0.5$.

④As $x \to +\infty$, $y \to \infty$ increasingly steeply.

⑤As $x \to -\infty$, $y \to 0$ from above but never reaches it.

Plot a couple of negative $x$-values. They show the asymptotic tail clearly and stop you drawing the curve stopping at the $y$-axis.

Worked Example 2 — A decay curve

Sketch $y = 100 \times 0.5^x$ and find when $y$ falls below $10$.

①$y$-intercept $(0, 100)$; $a = 0.5 < 1$, so this is decay.

②Values: $x=1 \to 50$; $x=2 \to 25$; $x=3 \to 12.5$; $x=4 \to 6.25$.

③So $y$ first falls below $10$ between $x = 3$ and $x = 4$.

④Exactly: $100(0.5)^x = 10 \Rightarrow 0.5^x = 0.1 \Rightarrow x = \dfrac{\log 0.1}{\log 0.5} = 3.32$

⑤Each unit increase in $x$ halves the value — so this function has a "half-life" of exactly $1$.

3 Finding $k$ and $a$ from Information

The value at $x = 0$ gives $k$ directly, since $a^0 = 1$.
Substitute a second point to get an equation in $a$.
Solve for $a$, taking an appropriate root.
Check both points satisfy your final formula.

Worked Example 3 — Two points

A curve $y = ka^x$ passes through $(0, 5)$ and $(3, 40)$. Find $k$ and $a$.

①At $x=0$: $y = k$, so $k = 5$.

②At $x=3$: $5a^3 = 40$, so $a^3 = 8$.

③$a = 2$

④$y = 5 \times 2^x$. Check: at $x=3$, $5 \times 8 = 40$ ✓

Worked Example 4 — Neither point at zero

$y = ka^x$ passes through $(1, 12)$ and $(4, 324)$. Find $k$ and $a$.

①$ka = 12$ and $ka^4 = 324$

②Divide the second by the first to eliminate $k$:

③$\dfrac{ka^4}{ka} = a^3 = \dfrac{324}{12} = 27$

④$a = 3$

⑤Then $3k = 12$, so $k = 4$, and $y = 4 \times 3^x$.

⑥Check at $x=4$: $4 \times 81 = 324$ ✓

Dividing the two equations is the key move. It removes $k$ in one step, leaving a simple power equation in $a$. Subtracting would not work, because $k$ is a factor rather than a term.

4 Related Forms

Two forms worth recognising

$y = a^{-x} = \left(\dfrac1a\right)^x$ — decay; the reflection of $a^x$ in the $y$-axis
$y = ka^x + c$ — shifted up by $c$, so the asymptote becomes $y = c$

Worked Example 5 — A shifted asymptote

Describe the graph of $y = 20 \times 0.7^x + 15$, including its asymptote, and interpret it as a cooling model.

①At $x = 0$: $y = 20 + 15 = 35$.

②As $x \to \infty$, $0.7^x \to 0$, so $y \to 15$.

③The asymptote is $y = 15$, not $y = 0$.

④As a model: an object starting at $35$ °C cooling towards a room temperature of $15$ °C.

⑤The $20$ is the initial excess temperature above the room, and it is that excess which decays by $30\%$ each period.

In a cooling model the exponential applies to the difference, not the temperature. Reading the $15$ as "the temperature falls to $15$ and stops" is right; reading it as part of the decaying term is not.

5 Quick Reference

The form

$y = ka^x$, variable in the index.

$k$

The value at $x = 0$.

$a$

The multiplier per unit step.

$a > 1$

Growth.

$0 < a < 1$

Decay.

Asymptote

$y = 0$, never reached.

No $x$-intercept

$ka^x$ is always positive.

Two points

Divide the equations to eliminate $k$.

$a^{-x}$

Reflection in the $y$-axis.

$+c$

Moves the asymptote to $y = c$.

6 Practice Questions

Question 1

Write down the $y$-intercept of $y = 9 \times 5^x$.

▶ Show solution

$(0, 9)$, since $5^0 = 1$.

Question 2

State whether each shows growth or decay: (a) $y = 3 \times 1.4^x$, (b) $y = 50 \times 0.9^x$.

▶ Show solution

(a) $a = 1.4 > 1$, so growth.

(b) $a = 0.9 < 1$, so decay.

Question 3

Find $y$ when $x = 3$ for $y = 6 \times 2^x$.

▶ Show solution

$y = 6 \times 8 = 48$

Question 4

Find $y$ when $x = -2$ for $y = 20 \times 3^x$.

▶ Show solution

$3^{-2} = \dfrac19$

$y = \dfrac{20}{9} = 2.22$ (3 s.f.)

Question 5

A curve $y = ka^x$ passes through $(0, 7)$ and $(2, 63)$. Find $k$ and $a$.

▶ Show solution

$k = 7$ from the first point.

$7a^2 = 63$, so $a^2 = 9$ and $a = 3$ (taking the positive root).

$y = 7 \times 3^x$

Question 6

A curve $y = ka^x$ passes through $(1, 6)$ and $(3, 54)$. Find $k$ and $a$.

▶ Show solution

$ka = 6$ and $ka^3 = 54$.

Dividing: $a^2 = 9$, so $a = 3$.

Then $3k = 6$, giving $k = 2$, so $y = 2 \times 3^x$.

Check: at $x=3$, $2 \times 27 = 54$ ✓

Question 7

State the equation of the asymptote of $y = 8 \times 0.5^x + 3$.

▶ Show solution

As $x \to \infty$, $0.5^x \to 0$, so $y \to 3$.

The asymptote is $y = 3$.

Question 8

Explain why $y = 5 \times 2^x$ never equals zero.

▶ Show solution

$2^x$ is positive for every real $x$: for large negative $x$ it becomes very small, but a positive number raised to any real power stays positive.

Multiplying by the positive $5$ keeps it positive, so $y > 0$ always and the curve never meets the $x$-axis.

Question 9

Compare $y = 100x$ and $y = 2^x$ at $x = 5$, $10$ and $20$, and comment.

▶ Show solution

$x$	$100x$	$2^x$
$5$	$500$	$32$
$10$	$1000$	$1024$
$20$	$2000$	$1\,048\,576$

The linear function is far ahead at $x=5$, they are almost level at $x=10$, and by $x=20$ the exponential is over five hundred times larger.

An exponential with $a > 1$ always overtakes any linear function eventually, no matter how large the linear gradient.

Question 10

A colony of bacteria is modelled by $N = 400 \times 1.3^t$, where $t$ is in hours.

(a) State the initial population and the hourly growth rate as a percentage. (b) Find the population after $6$ hours. (c) Find how long it takes the population to double, to the nearest minute. (d) The dish can hold $50\,000$ bacteria. Find when the model predicts it becomes full, and comment on the model's realism.

▶ Show solution

(a) At $t = 0$: $N = 400 \times 1 = 400$ bacteria.

The multiplier $1.3$ means a $\mathbf{30\%}$ increase each hour.

(b) $N = 400 \times 1.3^6 = 400 \times 4.826809 = 1930.7$

About $\mathbf{1931}$ bacteria (a whole number, since bacteria are counted).

(c) Doubling means $N = 800$:

$400 \times 1.3^t = 800 \;\Rightarrow\; 1.3^t = 2$

$t = \dfrac{\log 2}{\log 1.3} = \dfrac{0.301030}{0.113943} = 2.6419$ hours

$0.6419 \times 60 = 38.5$ minutes, so about $\mathbf{2}$ hours $39$ minutes.

Note the doubling time is the same wherever you start — from $400$ to $800$, or $5000$ to $10\,000$. That is characteristic of exponential growth.

(d) $400 \times 1.3^t = 50\,000 \;\Rightarrow\; 1.3^t = 125$

$t = \dfrac{\log 125}{\log 1.3} = \dfrac{2.096910}{0.113943} = 18.4$ hours

Comment: the model is likely to be good early on, while food and space are plentiful, but unrealistic as the dish fills. Real populations slow as resources run short, so growth tapers off well before the hard limit is reached.

In practice the true population would level off towards $50\,000$ rather than hitting it abruptly at $18.4$ hours — so that figure should be read as "the model breaks down by about then", not as a prediction.

Exponential Functions (EL1) · OCR FSMQ Additional Maths · Created with MathJax