1A - Exponential functions

An exponential function changes by a constant multiplier over equal input steps. You will build its values, recognise growth and decay graphs, and read the effect of multiplying or shifting an exponential expression.

Multiplicative change

A quantity multiplies by $4$ whenever its input increases by $1$ . What multiplier should it use over a half-step?

Two half-steps make one whole step. If each half-step uses multiplier $r>0$ , then $r^2=4$ , so $r=2$ . Adding half of $4$ is not the reason: multiplying twice by $2$ gives the required factor $4$ .

This is the index-law meaning of $4^{\frac12}=2$ . It lets a pattern of repeated multiplication extend beyond whole-number inputs.

Exponential function

An exponential function has the input in the exponent. Its basic form is $f(x)=a^x$ , where the constant base satisfies $a>0$ and $a\ne1$ .

For example, $2^x$ is exponential. The function $x^2$ is a power function: its variable is the base instead. Base $1$ gives the constant function $1^x=1$ , which we exclude from growth and decay.

Integer, negative and fractional indices give exact points. The exponential function extends this pattern smoothly to every real input; a calculator can approximate values at other inputs too.

Equal steps give equal multipliers

For $f(x)=a^x$ , with $a>0$ and $a\ne1$ ,

f(x+h)=a^{x+h}=a^x a^h.

So an input step of $h$ multiplies the output by $a^h$ . A unit step uses multiplier $a$ .

Not on the formulae sheet: learn it.

Build a table using the index rules, rather than guessing a straight-line pattern between entries.

Worked example 1: unit input steps

Question. Find the values of $y=4^x$ at $x=-1,0,1,2$ . State the multiplier from one listed value to the next.

Plan. Evaluate the negative and zero powers first, then multiply by $4$ for each unit step.

4^{-1}=\frac14,\qquad 4^0=1,\qquad4^1=4,\qquad4^2=16.

$x$	$-1$	$0$	$1$	$2$
$4^x$	$\frac14$	$1$	$4$	$16$

Answer. The values are $\frac14,1,4,16$ ; each successive value is multiplied by $4$ .

Check. The adjacent ratios are $1\div\frac14=4$ , $4\div1=4$ and $16\div4=4$ . Their differences are unequal, so this is not constant additive change.

Keep the same function, but halve the input step. The multiplier must change as well.

Worked example 2: half-unit input steps

Question. Find the values of $y=4^x$ at $x=-\frac12,0,\frac12,1$ . State the multiplier from one listed value to the next.

Plan. Use $4^{\frac12}=2$ and its reciprocal, then check the ratios.

4^{-\frac12}=\frac1{\sqrt4}=\frac12, \qquad4^0=1, \qquad4^{\frac12}=2, \qquad4^1=4.

$x$	$-\frac12$	$0$	$\frac12$	$1$
$4^x$	$\frac12$	$1$	$2$	$4$

Answer. The values are $\frac12,1,2,4$ ; each successive value is multiplied by $2$ .

Check. Two half-steps multiply by $2\times2=4$ , agreeing with the unit-step multiplier in Example 1.

The multiplier belongs to a specified input step. You cannot compare ratios from unequal steps as if the steps were equal.

Watch out

For $4^x$ , halfway between inputs $0$ and $1$ gives output $2$ , not the arithmetic midpoint $\frac{1+4}{2}=2.5$ . Exponential outputs multiply over equal steps; a straight line's outputs have equal differences.

Try half-steps with a different square-number base.

Exponential graph shapes

The base tells you whether the graph rises or falls, while positive outputs and the zero-index rule give features shared by both shapes.

For $y=2^x$ , each unit step doubles the output, so the graph increases. In the negative direction the values halve: $1,\frac12,\frac14,\frac18,\ldots$ . They get as close to $0$ as you wish, but remain positive.

For $y=\left(\frac12\right)^x=2^{-x}$ , each unit step halves the output, so the graph decreases. Replacing $x$ by $-x$ swaps the left and right sides of the graph of $2^x$ : this is reflection in the $y$ -axis.

[DIAGRAM: asset_name: Growing and decaying exponentials; asset_slug: exponential-shapes; caption: Positive exponential graphs rise for a base above 1 and fall for a base between 0 and 1; both pass through (0, 1) and approach y = 0; description: Two coordinate plots on equal scales show y equals 2 to the x and y equals one half to the x. Each has the point (0, 1) marked. The increasing curve approaches the horizontal line y equals 0 towards the left; the decreasing curve approaches that line towards the right. The curves are reflections in the y-axis and remain strictly above the x-axis.]
Positive exponential graphs rise for a base above 1 and fall for a base between 0 and 1

The graph is a smooth curve through its points. Joining a few calculated points with straight line segments would hide the continuous multiplicative change.

Domain, range and horizontal asymptote

The domain is the set of allowed inputs; the range is the set of resulting outputs. A horizontal asymptote is a horizontal line that the curve approaches as the input becomes very large in the positive or negative direction.

For $y=a^x$ , with $a>0$ and $a\ne1$ , the domain is all real $x$ , the range is $y>0$ , the $y$ -intercept is $(0,1)$ , and the horizontal asymptote is $y=0$ . This curve never meets its asymptote, so it has no $x$ -intercept.

A base greater than $1$ gives exponential growth: the output increases as $x$ increases. A base between $0$ and $1$ gives exponential decay: the output decreases as $x$ increases. Decay still has positive outputs; it does not mean the graph lies below the axis.

Worked example 3: a growth sketch

Question. Describe a sketch of $y=2^x$ , giving three exact points, its asymptote, domain and range.

Plan. Use $x=-1,0,1$ for points, then use the multiplier and positive values to decide the shape and its end behaviour.

\begin{aligned} 2^{-1}&=\frac12 &&\text{a negative index gives a reciprocal}\\ 2^0&=1 &&\text{locate the }y\text{-intercept}\\ 2^1&=2 &&\text{confirm the growth direction} \end{aligned}

Draw a smooth increasing curve through $(-1,\frac12)$ , $(0,1)$ and $(1,2)$ . Towards the left it approaches $y=0$ from above; towards the right its values grow without bound, as in the left-hand graph.

Answer. An increasing curve with horizontal asymptote $y=0$ , domain all real $x$ and range $y>0$ .

Check. Every unit step doubles the output, including $\frac12$ to $1$ to $2$ . No plotted point is on or below the asymptote.

Now replace the base $2$ by $\frac12$ . The intercept and range stay the same, but the direction and the end approaching zero swap.

Worked example 4: a decay sketch

Question. Describe a sketch of $y=\left(\frac12\right)^x$ , giving three exact points, its asymptote, domain and range.

Plan. Evaluate at the same inputs and use the half-size multiplier to choose the decreasing shape.

\begin{aligned} \left(\frac12\right)^{-1}&=2 &&\text{take the reciprocal}\\ \left(\frac12\right)^0&=1 &&\text{the zero power is unchanged}\\ \left(\frac12\right)^1&=\frac12 &&\text{one step halves the output} \end{aligned}

Draw a smooth decreasing curve through $(-1,2)$ , $(0,1)$ and $(1,\frac12)$ . It approaches $y=0$ from above towards the right, as in the right-hand graph; towards the left its values grow without bound.

Answer. A decreasing curve with horizontal asymptote $y=0$ , domain all real $x$ and range $y>0$ .

Check. Its values at opposite inputs match those of $2^x$ , since $\left(\frac12\right)^x=2^{-x}$ . This confirms the reflection in the $y$ -axis.

A sketch should show its defining features clearly, even when it has no detailed grid.

Exam tip

Label the axes, $y$ -intercept and asymptote. Write the asymptote as the equation $y=0$ , and let the curve approach it smoothly without touching it. A decay curve passes through $(0,1)$ , not the origin.

Use the base to identify the next curve without building a long table.

Interpreting exponential models

A multiplier outside the power changes the scale of the outputs, while an added constant moves the graph's horizontal asymptote.

For example, $N(t)=120(1.1)^t$ models a quantity that starts at $120$ when $t=0$ and multiplies by $1.1$ each unit of time: a $10\%$ increase. A multiplier $0.8$ would retain $80\%$ each step, a $20\%$ decrease. A time model may use only $t\ge0$ , even though the underlying exponential function allows every real input.

For shifted graphs, work from the expression itself. In $2^{x-1}$ , the exponent becomes zero at $x=1$ , so the point $(0,1)$ on $2^x$ moves right to $(1,1)$ . Multiplying by $3$ moves that point to $(1,3)$ ; adding $2$ moves it to $(1,5)$ . A negative multiplier reverses the vertical direction.

Reading a shifted exponential

For $y=Aa^{x-h}+k$ , where $a>0$ , $a\ne1$ and $A\ne0$ :

the graph of $a^x$ is shifted right by $h$ , its heights are multiplied by $A$ , then it is shifted up by $k$ ;
the horizontal asymptote is $y=k$ ;
the point at $x=h$ is $(h,A+k)$ ;
the $y$ -intercept is $(0,Aa^{-h}+k)$ , found by substituting $x=0$ .

The domain is all real $x$ . The range is $y>k$ if $A>0$ , and $y<k$ if $A<0$ , because $a^{x-h}$ is always positive. The graph never reaches $y=k$ .

Not on the formulae sheet: learn it.

A negative $h$ shifts left; a negative $k$ shifts down. When $A<0$ , multiplying the exponential part by a negative number reflects its heights before the final vertical shift. The sign of $A$ reverses increasing and decreasing behaviour.

[DIAGRAM: asset_name: Reflecting a shifted exponential; asset_slug: shifted-exponentials; caption: Changing the coefficient from 3 to minus 3 reflects the shifted exponential in the line y = 2; description: A shared coordinate plot shows y equals 3 times 2 to the power x minus 1 plus 2 and y equals minus 3 times 2 to the power x minus 1 plus 2. Both approach the dashed horizontal line y equals 2. The points (1, 5) and (1, minus 1) are marked at equal vertical distances above and below that line. The positive-coefficient curve increases and stays above 2; the negative-coefficient curve decreases and stays below 2.]
Changing the coefficient from 3 to minus 3 reflects the shifted exponential in the line y = 2

The two curves in the picture have the same base and shifts. Changing only the sign of the coefficient makes their distances from $y=2$ equal and opposite.

Worked example 5: a scale and two shifts

Question. For $y=3(2^{x-1})+2$ , find the $y$ -intercept, horizontal asymptote and range. State whether the function increases or decreases.

Plan. Substitute $x=0$ , then read the vertical shift and the sign of the exponential part.

\begin{aligned} y(0)&=3(2^{-1})+2 &&\text{put }x=0\\ &=\frac32+2=\frac72 &&\text{use the reciprocal power}\\ y-2&=3(2^{x-1})>0 &&\text{the exponential part is positive} \end{aligned}

As $x$ becomes very negative, $2^{x-1}$ approaches zero, so $y$ approaches $2$ . The base exceeds $1$ and the coefficient is positive, so the function increases.

Answer. $y$ -intercept $(0,\frac72)$ ; horizontal asymptote $y=2$ ; range $y>2$ ; increasing.

Check. At $x=1$ , the value is $3(2^0)+2=5$ . At $x=2$ , it is $8$ . The distances from the asymptote are $3$ and $6$ : they double, as required by the base $2$ .

Change only the coefficient from $3$ to $-3$ . The asymptote stays in place, while the side of the asymptote and the direction of change reverse.

Worked example 6: a negative coefficient

Question. For $y=-3(2^{x-1})+2$ , find the $y$ -intercept, horizontal asymptote and range. State whether the function increases or decreases.

Plan. Substitute $x=0$ , then use the negative coefficient to interpret the positive exponential factor.

\begin{aligned} y(0)&=-3(2^{-1})+2 &&\text{put }x=0\\ &=-\frac32+2=\frac12 &&\text{keep the negative coefficient}\\ y-2&=-3(2^{x-1})<0 &&\text{negative times positive is negative} \end{aligned}

The exponential factor grows as $x$ increases, so its negative multiple becomes more negative. The graph decreases and approaches $y=2$ from below towards the left.

Answer. $y$ -intercept $(0,\frac12)$ ; horizontal asymptote $y=2$ ; range $y<2$ ; decreasing.

Check. At $x=1$ , the value is $-1$ ; at $x=2$ , it is $-4$ . Their distances below $2$ are $3$ and $6$ . Also, the two worked-example outputs at any common input add to $4$ , confirming reflection in $y=2$ .

With a vertical shift, the output itself no longer has a constant ratio: its distance from the asymptote does. In Example 5, the outputs $5$ and $8$ do not have ratio $2$ , but $5-2=3$ and $8-2=6$ do.

Watch out

The coefficient $A$ is not automatically the $y$ -intercept of $Aa^{x-h}+k$ . Substituting $x=0$ accounts for both shifts. Likewise, exponential decay from a base below $1$ stays positive before any negative coefficient or vertical shift is applied.

The opening half-step multiplier is $2$ , because $4^{\frac12}=2$ . A scale or vertical shift can move the graph, but the same multiplier rule still controls equal input steps in its exponential part.

Quick Recap

Equal input steps in $a^x$ give equal multipliers; a step of $h$ uses $a^h$ .
The basic graph passes through $(0,1)$ , has range $y>0$ and approaches $y=0$ .
A base above $1$ gives growth; a base between $0$ and $1$ gives decay.
For $Aa^{x-h}+k$ , substitute zero for the intercept and use $y=k$ for the asymptote. The sign of $A$ determines which side of that line the graph occupies.

A half-step in $4^x$ multiplies its value by $2$ ; two such steps multiply by $4$ .

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1A - Exponential functions

Multiplicative change

Exponential graph shapes

Interpreting exponential models

Quick Recap

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