4.04 Graphing techniques using calculus

The first and second derivatives of a function have been shown to give us a wealth of information about the shape of the curve. Combining this with our previous understanding of functions, specifically symmetry, domain, and range, we are able to create a checklist of skills that allow us to sketch any function that is not easily recognisable.

Sketching functions

Strategy	Description
Determine any special characteristics	Investigate values that $f(x)$f(x) can't take, such as within fractions or square root functions. Check if there are any asymptotes or discontinuities.
Symmetry, even/odd functions	Check for an even function $f(x)=f(-x)$f(x)=f(−x) Check for an odd function $f(-x)=-f(x)$f(−x)=−f(x)
Domain and range	Consider the $x$x and $y$y-values that a function can take. If a domain is specified, clearly indicate the coordinates of the end points of your function sketch.
Find axis intercepts	Make $y=0$y=0 to find any $x$x-intercepts, if possible. Make $x=0$x=0 to find the $y$y-intercept.
Determine stationary points	Consider $\frac{dy}{dx}=0$dydx​=0
Determine the nature of stationary points	Consider $\frac{dy}{dx}$dydx​ left and right of each stationary point Or Consider the sign of $\frac{d^2y}{dx^2}$d2ydx2​ at each stationary point
Find any possible points of inflection	Consider $\frac{d^2y}{dx^2}=0$d2ydx2​=0 and test for a change in concavity

Remember, when sketching functions, use a third of an $A_4$A4​ page and adjust the scales to suit. There is no need for the $x$x and $y$y-axis scales to exactly match each other, or to be marked in regular intervals. Coordinates of all the important points for your function (stationary points, intercepts, points of inflection and limits of the domain if applicable) should be indicated either by marking the coordinates at the point or clearly on each axis.

Worked examples

Example 1

Sketch the graph of the function $y=\frac{x^2}{3}-x+\frac{2}{3}$y=x23​−x+23​ in the domain $[-3,3]$[−3,3].

Think: This is a quadratic function with a specified domain. As the leading coefficient is positive it will be concave upwards and have a minimum turning point. We will confirm this using calculus.

Do: First we will find the intercepts and the function values at the end points of the domain.

To find the $y$y-intercept, make $x=0$x=0.

$y$y	$=$=	$\frac{x^2}{3}-x+\frac{2}{3}$x23​−x+23​
$y$y	$=$=	$\frac{0^2}{3}-0+\frac{2}{3}$023​−0+23​
$y$y	$=$=	$\frac{2}{3}$23​

 

Therefore, there is a $y$y intercept at $(0,\frac{2}{3})$(0,23​).

To find the $x$x-intercept/s, make $y=0$y=0.

$y$y	$=$=	$\frac{x^2}{3}-x+\frac{2}{3}$x23​−x+23​
$0$0	$=$=	$\frac{x^2}{3}-x+\frac{2}{3}$x23​−x+23​
$0$0	$=$=	$x^2-3x+2$x2−3x+2
$0$0	$=$=	$(x-2)(x-1)$(x−2)(x−1)
$x$x	$=$=	$2,1$2,1

 

Therefore, there are $x$x intercepts at $(2,0)$(2,0) and $(1,0)$(1,0).

To find the function values at the limits of the domain make $x=\pm3$x=±3.

For $x$x	$=$=	$3$3
$y$y	$=$=	$\frac{3^2}{3}-3+\frac{2}{3}$323​−3+23​
$y$y	$=$=	$\frac{2}{3}$23​
For $x$x	$=$=	$-3$−3
$y$y	$=$=	$\frac{(-3)^2}{3}-(-3)+\frac{2}{3}$(−3)23​−(−3)+23​
$y$y	$=$=	$\frac{20}{3}$203​

 

Therefore, the end points of our function sketch will be at $(3,\frac{2}{3})$(3,23​) and $(-3,\frac{20}{3})$(−3,203​).

To find any stationary points make $\frac{dy}{dx}=0$dydx​=0.

$\frac{dy}{dx}$dydx​	$=$=	$\frac{2x}{3}-1$2x3​−1
$0$0	$=$=	$\frac{2x}{3}-1$2x3​−1
$1$1	$=$=	$\frac{2x}{3}$2x3​
$3$3	$=$=	$2x$2x
$x$x	$=$=	$\frac{3}{2}$32​

 

To find the $y$y-coordinate of this stationary point and determine its nature we substitute $x=\frac{3}{2}$x=32​ into the original function and $\frac{d^2y}{dx^2}$d2ydx2​ respectively.

$y$y	$=$=	$\frac{(\frac{3}{2})^2}{3}-\frac{3}{2}+\frac{2}{3}$(32​)23​−32​+23​
$y$y	$=$=	$\frac{9}{12}-\frac{3}{2}+\frac{2}{3}$912​−32​+23​
$y$y	$=$=	$\frac{9-18+8}{12}$9−18+812​
$y$y	$=$=	$-\frac{1}{12}$−112​
$\frac{d^2y}{dx^2}$d2ydx2​	$=$=	$\frac{2}{3}$23​

 

Therefore, as $\frac{d^2y}{dx^2}$d2ydx2​ is always positive the curve is concave up and the stationary point at $\left(\frac{3}{2},-\frac{1}{12}\right)$(32​,−112​) is a minimum turning point.

We now have enough information to sketch this quadratic as follows, remembering to show all important points and features on the graph.

 

Global maximum and minimum points

You may be asked to determine the maximum or minimum value of a function for a given domain. Sometimes, this will be a turning point, but often the maximums and minimums are at the end points of the domain. A point is considered a global maximum or minimum if it is the biggest or smallest function value for the given domain. Turning points that don't meet this criteria are called local maximums and minimums.

For example, the turning point of a quadratic will always be either the global maximum or minimum.

A cubic, however, could have a local maximum or minimum point that is not necessarily the maximum or minimum value the function could take.

Turning points and points of inflection

When sketching functions it is important to make sure that any turning points and points of inflection are drawn as smooth curves, that is, without sharp corners. Additionally, horizontal points of inflection and regular points of inflection must reflect their different gradient properties. Notice the appearance of the two graphs below. The graph on the left has a horizontal point of inflection at $A$A. The graph on the right has a regular point of inflection at point $B$B.

 

Behaviour as $x\rightarrow\pm\infty$x→±∞

It is useful when sketching curves to consider what happens to a function as $x\rightarrow\pm\infty$x→±∞. This gives us an indication of what the graph is doing as $x$x gets very large ($x\rightarrow\infty$x→∞) and very small ($x\rightarrow-\infty$x→−∞). Here are some general suggestions for analysing a function as $x\rightarrow\pm\infty$x→±∞

• The limiting behaviour of polynomials as $x\rightarrow\pm\infty$x→±∞ is determined by the leading term, that is, the one with the highest index. For example, for the polynomial $P(x)=x^3+3x^2-10x$P(x)=x3+3x2−10x, as $x\rightarrow\pm\infty$x→±∞ the term $x^3$x3 dictates that the function will be either large and positive or large and negative depending on the sign of the input.

• The limiting behaviour of $\frac{1}{x}$1x​ as $x\rightarrow\pm\infty$x→±∞ is useful to understand for functions with the variable in the denominator such as hyperbolic functions. If we divide $1$1 by a number that gets larger and larger, we get a number very close to $0$0.

Let's look at the graph of the hyperbola $f(x)=\frac{1}{x}$f(x)=1x​ to highlight this:

Example 2

Sketch the function $f(x)=2x^3-9x^2+12x-4$f(x)=2x3−9x2+12x−4, showing all important points and features.

Think: This function is a cubic and could have up to $2$2 turning points. Using $f'(x)$f′(x) and $f''(x)$f′′(x) will be vital in determining all the important points and features of this graph. We will start by finding the $y$y-intercept.

Do: To find the $y$y intercept/s, make $x=0$x=0.

$f(0)$f(0)	$=$=	$2\times0^3-9\times0^2+12\times0-4$2×03−9×02+12×0−4
$f(0)$f(0)	$=$=	$-4$−4

 

Therefore, there is a $y$y-intercept at $(0,-4)$(0,−4).

To find any $x$x-intercepts may prove difficult so we will move straight onto finding (and determining) the nature of any stationary points by looking at the first and second derivatives.

 

$f'(x)$f′(x)	$=$=	$6x^2-18x+12$6x2−18x+12
$0$0	$=$=	$6x^2-18x+12$6x2−18x+12
$x^2-3x+2$x2−3x+2	$=$=	$0$0
$(x-1)(x-2)$(x−1)(x−2)	$=$=	$0$0
$x$x	$=$=	$1,2$1,2

 

Therefore, there are stationary points at $x=1,2$x=1,2.

To find the $y$y-coordinates of these stationary points we substitute them back into the original function.

For $x$x	$=$=	$1$1
$f(1)$f(1)	$=$=	$2\times1^3-9\times1^2+12\times1-4$2×13−9×12+12×1−4
$f(1)$f(1)	$=$=	$1$1
For $x$x	$=$=	$2$2
$f(2)$f(2)	$=$=	$2\times2^3-9\times2^2+12\times2-4$2×23−9×22+12×2−4
$f(2)$f(2)	$=$=	$0$0

 

Therefore the coordinates of the stationary points are $(1,1)$(1,1) and $(2,0)$(2,0) respectively.

To determine the nature of these points we could look at the sign of the first derivative left and right of each point:

$x$x	$0$0	$1$1	$1.5$1.5	$2$2	$3$3
$f'(x)$f′(x)	$12$12	$0$0	$-1.5$−1.5	$0$0	$12$12
Gradient	/	.	\	.	/
Turning point classification		local maximum		local minimum

Or we can investigate the sign of $f''(x)$f′′(x). Let's look at the second derivative at each of the points.

$f'(x)$f′(x)	$=$=	$6x^2-18x+12$6x2−18x+12
$f''(x)$f′′(x)	$=$=	$12x-18$12x−18

 

We can see that $f''(x)=0$f′′(x)=0 at $x=\frac{3}{2}$x=32​. This is a possible point of inflection. We will create a table and include this point to determine the nature of this and the stationary points:

$x$x	$1$1	$\frac{3}{2}$32​	$2$2
$f''(x)$f′′(x)	$-6$−6	$0$0	$6$6
Concavity	negative	zero	positive
Meaning	concave down	no concavity	concave up
Classification	local maximum	point of inflection	local minimum

 

Therefore, as $f'(x)=0$f′(x)=0 and $f''(x)$f′′(x) is negative at $x=1$x=1 this point is a local maximum.

Therefore, as $f''(x)=0$f′′(x)=0 and the concavity changes from negative to positive there is a point of inflection at $x=\frac{3}{2}$x=32​.

Therefore, as $f'(x)=0$f′(x)=0 and $f''(x)$f′′(x) is positive at $x=2$x=2 this point is a local minimum.

The last thing we will do is look at the behaviour of the function as $x\rightarrow\pm\infty$x→±∞. As this is a cubic, we can quickly see, looking at the leading term $2x^3$2x3 the function value will be large and positive for $x\rightarrow\infty$x→∞ and large and negative for $x\rightarrow-\infty$x→−∞.

Lets now sketch our function using all the information we have determined and showing all the important points:

Practice questions

Question 1

Question 2

Question 3