Exponential and log graphs

Intercepts, asymptotes, domain, sign, range

In 4 of 20 recent exams
Paper 1, no calculator
1 to 4 marks

How it is set in the exam

Paper 1 (no calculator) only, 1–4 marks a part: usually the opening parts of a longer question that goes on to differentiate (a maximum point, a sketch); a sign part can also come later, just before the sketch. Typical (4 of 20 exam versions, 8 parts).

Spot it in the exam

Exam phrases to spot (6)

“find the coordinates of the points where the graph of $f$ meets the axes”
“write down the equation of the vertical asymptote”
“write down the equation of the horizontal asymptote”
“state the largest possible domain of $f$ ”
“find the values of $x$ for which $f(x) > 0$ ”
“the graph of $f$ has a maximum point at …; write down the range of $f$ ”

In disguise (4)

a sign part inside a calculus question on $\frac{\ln x}{x^n}$ , even after $f'$ has been found (for example just before the sketch): the sign comes from $e^{\ldots} > 0$ and the sign of $\ln$ (positive exactly when its argument is more than $1$ ), not from $f'$
"explain why $f$ has no zeros": $e^{\ldots} > 0$ , so a sum of positive multiples of exponentials is never $0$
"what happens to $f(x)$ for large negative $x$ ": an asymptote in words
the vertical asymptotes of a log of a quadratic, after an earlier part found its domain: the quadratic is $0$ at its roots, which are the ends of that domain, so each root gives an asymptote $x = \ldots$ , written down with no new working

Variants (4)

an exponential expression with no constant term (a difference of powers of $e^x$ , a product or a quotient): both intercepts exactly, by factoring out the lowest power
$\ln$ of a linear expression: its domain and asymptote (its zero, argument $= 1$ , is one line: the $\log_a X = c$ row of Solve a logarithmic equation)
a $\ln$ over a power of $x$ , or over $e^x$ : where it is positive
the range, when the question gives the maximum or minimum

Looks like it, but isn't (12)

the zeros or $x$ -intercepts of a log function: argument $= 1$ is an equation to solve (one line for a linear argument, a quadratic for a quadratic one), Solve a logarithmic equation. Only inside a question on the log's domain and asymptote is that one line a step here (Example 1.2(c))
the domain of a log of a quadratic: Quadratic inequalities
the greatest value of a log of a quadratic: at the quadratic's vertex, Find the vertex
a quadratic in $e^x$ with a constant term: Quadratic in disguise
a log function evaluated at a number: Exact value of a logarithm
a $y$ -intercept by plain substitution, with no exponential or log fact: Key features of a graph
a quantity changing over time in a context: Exponential and log models; the value such a model tends to in the long run, found as a limit: Rate of change in context
the asymptotes of $\frac{ax + b}{cx + d}$ : Rational functions
a turning point that is not given and has to be found with $f'$ : Stationary points
the formula of an inverse: Find the inverse function
the features of the image of an exponential or log graph under a stated transformation: Transform a graph
where a graph is above or below a level other than $0$ , read off a sketch or found on the GDC ("hence write down the values of $x$ for which $f(x) > 2$ "): Solve graphically

Worked example

An exponential expression: both intercepts, the asymptote, the range

Paper 1 (no calculator)

Question

Let $f(x) = 6e^{2x} - e^{3x}$ , for $x \in \mathbb{R}$ .
(a) Find the coordinates of the point where the graph of $f$ crosses the $y$ -axis. [2]
(b) Find the coordinates of the point where the graph of $f$ crosses the $x$ -axis. [4]
(c) Write down the equation of the horizontal asymptote of the graph of $f$ . [1]
(d) The graph of $f$ has a maximum point at $(\ln 4, 32)$ . Write down the range of $f$ . [1]

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Which method?

An $x$ -intercept means an expression equal to $0$ . Its form decides the first move:

no constant term: every term is a multiple of a power of $e^x$ , or of $a^x$ ( $6e^{2x} - e^{3x}$ )
First move
factor out the lowest power, and reject it: $e^{2x} > 0$
Where it is taught
here
one power of $e^x$ and a constant ( $7 - 3e^{-x}$ )
First move
isolate the power, then take $\ln$
Where it is taught
Solve an exponential equation
a constant, and powers in the ratio $2 : 1$ ( $e^{2x} - 5e^x + 6$ )
First move
$u = e^x$ gives a quadratic; reject $u \le 0$
Where it is taught
Quadratic in disguise
$\ln$ of a linear expression ( $\ln(6 - 2x)$ )
First move
the argument $= 1$ : one line
Where it is taught
Solve a logarithmic equation (the $\log_a X = c$ row; a step here only inside a domain-and-asymptote question, Example 1.2(c))
$\ln$ of a quadratic ( $\ln(x^2 - 5x + 5)$ )
First move
the argument $= 1$ : solve the quadratic; both roots are in the log's own domain (the argument is $1 > 0$ there), so only a domain the question states can reject one
Where it is taught
Solve a logarithmic equation

You see	First move	Where it is taught
no constant term: every term is a multiple of a power of $e^x$ , or of $a^x$ ( $6e^{2x} - e^{3x}$ )	factor out the lowest power, and reject it: $e^{2x} > 0$	here
one power of $e^x$ and a constant ( $7 - 3e^{-x}$ )	isolate the power, then take $\ln$	Solve an exponential equation
a constant, and powers in the ratio $2 : 1$ ( $e^{2x} - 5e^x + 6$ )	$u = e^x$ gives a quadratic; reject $u \le 0$	Quadratic in disguise
$\ln$ of a linear expression ( $\ln(6 - 2x)$ )	the argument $= 1$ : one line	Solve a logarithmic equation (the $\log_a X = c$ row; a step here only inside a domain-and-asymptote question, Example 1.2(c))
$\ln$ of a quadratic ( $\ln(x^2 - 5x + 5)$ )	the argument $= 1$ : solve the quadratic; both roots are in the log's own domain (the argument is $1 > 0$ there), so only a domain the question states can reject one	Solve a logarithmic equation

In the first row, taking $\ln$ of both sides also scores ( $6e^{2x} = e^{3x}$ gives $\ln 6 + 2x = 3x$ , so $x = \ln 6$ ); with a base such as $2$ , writing both terms as powers of it and equating the powers does the same. Substituting $u = e^x$ works too, but it gives a cubic, not a quadratic: $6u^2 - u^3 = u^2(6 - u) = 0$ , and then $u = 0$ must be rejected, because $e^x > 0$ . As an equation with no graph ("solve $6e^{2x} = e^{3x}$ "), the same move is Solve an exponential equation.

Steps

$y$ -intercept:
put $x = 0$ ( $e^0 = 1$ , $\ln 1 = 0$ ) and write the point, $(0, y)$ . If $0$ is outside the domain (a log), there is none.
$x$ -intercept:
set $f(x) = 0$ and use the table. For an exponential expression, factor out the power of $e^x$ and say why it can't be $0$ . A quotient is $0$ exactly where its top is $0$ and its bottom is not: a bottom such as $e^x + 3$ is never $0$ (it is more than $3$ ), so set the top to $0$ . For $\ln(\text{linear})$ , the argument $= 1$ (one line: the $\log_a X = c$ row of Solve a logarithmic equation). Write the point, $(x, 0)$ .
Asymptotes, as equations:
$c + b\,e^{kx}$ or $c + b\,a^x$ has $y = c$ on the side where the power tends to $0$ ; $\ln\left(g(x)\right)$ has $x = \ldots$ where $g(x) = 0$ . Check both ends: one end of an exponential expression can run off to $\pm\infty$ with no asymptote.
Domain:
a log's argument $> 0$ (flip the inequality if you divide by a negative). An exponential's domain is every $x$ .
Sign:
split $f$ into factors whose signs you know: $e^{\ldots} > 0$ , a square $> 0$ except where it is $0$ , and $\ln(g) > 0$ exactly when $g > 1$ . When only one factor can change sign and every other factor is positive, $f$ has that factor's sign. A factor that is always negative, such as a negative constant or $-e^x$ , reverses it: $-\frac{2\ln x}{x^2} < 0$ for $x > 1$ . If two factors can change sign, make a sign table. Exclude any $x$ where $f$ is undefined.
Range:
from the maximum or minimum (given, found in an earlier part, or the value at the end of a restricted domain such as $x \ge 0$ ) and the behaviour at both ends: $\le$ or $\ge$ at a value that is reached, $<$ or $>$ at an asymptote that is never reached.

Common mistakes

An intercept as one number
" $5$ " or " $x = \ln 6$ " when coordinates are asked scores . Write $(0, 5)$ and $(\ln 6, 0)$ .
An asymptote as a number
it is a line, so write $y = 0$ or $x = 3$ .
Dividing by $e^{2x}$ without saying why that is allowed.
It is allowed because $e^{2x} \ne 0$ ; say so. Dividing by $x$ is not allowed in the same way: $xe^x = 3x$ loses the root $x = 0$ .

More mistakes (5)

The log's domain the wrong way round
$\ln(6 - 2x)$ needs $x < 3$ . Dividing an inequality by a negative number flips it.
$\ln$ changes sign at $0$
it changes sign where its argument is $1$ . $\ln(6 - 2x) > 0$ means $6 - 2x > 1$ .
A range with the wrong ends
a maximum that is reached gives $\le$ ; an asymptote that is never reached gives $<$ . And check both ends: $6e^{2x} - e^{3x}$ has an asymptote on one side only.
A range written in $x$
" $x \le 32$ " describes $x$ -values, not values of $f$ . Markschemes accept $f(x) \le 32$ , $y \le 32$ or $(-\infty, 32]$ .
Forgetting a value where the function is undefined
in a sign answer: $\frac{\ln x}{(x - 2)^2}$ is positive for $x > 1$ except at $x = 2$ , where the bottom is $0$ .

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