How to Sketch a Graph Using the Derivative

Expert reviewed • 22 November 2024 • 8 minute read

HSC Maths Advanced Syllabus

use calculus to determine and verify the nature of stationary points, find local and global maxima and minima and points of inflection (horizontal or otherwise), examine behaviour of a function as 𝑥 → ∞ and 𝑥 → −∞ and hence sketch the graph of the function
- use calculus to establish the location of local and global maxima and minima, including checking endpoints of an interval if required
- evaluate solutions and their reasonableness given the constraints of the domain and formulate appropriate conclusions to optimisation problems

Note:

Video coming soon!

Steps to Follow

In the previous module, we learnt some curve sketching instructions to ensure efficiency when creating a graph. Now, after learning about the use of calculus when curve sketching, we can add a few more steps to ensure efficiency when sketching harder graphs.

Simplification and factorisation: It is important to first simplify any equation you have been asked to graph, to ensure that the following steps are easier.
Find intercepts: The next step is to find your $x$ and $y$ intercepts.
Find signs: once your intercepts have been found, it is important to determine the the nature of the curve between points/intercepts. By doing so, you will start to understand the general shape of the graph. Remember, the signs can be found by using a table of signs.
Determine horizontal and vertical asymptotes: Find the discontinuities and plot them on a cartesian plane using a dotted line
Use the first derivative: Find the stationary points and determine the gradient of the graph around these points, using a table of values.
Use the second derivative: Find the points of inflection and test the concavity of the stationary points.
Sketch the graph: Finally by following your table of signs, using your $x$ and $y$ intercepts, turning points, points of inflection and following the asymptotes, plot the graph. (Ensure all important aspects of the graph are labelled)

Practice Question 1

Graph the function $f(x)=\frac{x}{x^2-4}$

Upon visual inspection of the function we can see it is in it’s simplest form. Thus, the first step is to find the functions intercepts/zeroes.

0=\frac{x}{x^2-4}\\x=0

$\therefore$ The x-intercept of the graph is at $x=0$

y=\frac{0}{0^2-4}\\y=0

$\therefore$ The y-intercept of the graph is at $y=0$

Finding the vertical asymptote/s of the graph:

0=x^2-4\\x^2=4\\x=\pm2

$\therefore$ The vertical asymptotes of this graph are found when $x=2$ and $x=-2$

Finding the horizontal asymptotes of the graph:

Because the degree of $x$ is larger in the denominator, a horizontal asymptote occurs at $y=0$

Finding the stationary points:

Using the quotient rule to derive the function, we find the derivative to be…

f^`(x)=-\frac{x^2+4}{(x^2-4)^2}

Equating with zero:

0=-\frac{x^2+4}{(x^2-4)^2}\\0=x^2+4\\x^2=-4

Because it is not possible to square root a negative number, we can determine that this function has no stationary points.

Finding the points of inflection and testing concavity:

Using the quotient rule again to find the second derivative, we find…

f^{``}(x)=\frac{2x(x^2+12)}{(x^2-4)^3}

Equating with zero:

0=\frac{2x(x^2+12)}{(x^2-4)^3}\\0=2x(x^2+12)\\2x=0\quad , \quad x^2+12=0\\x=0\quad , \quad x^2=-12

Because it is not possible to square root a negative number, we can determine that this function has only one point of inflection when $x=0$ . But to find the point we must substitute this back into $f(x)$

f(0)=\frac{0}{0^2-4}\\=0

$\therefore$ There is a point of inflection at $(0,0)$

Now by substituting points to find the general shape of the graph, and implementing all points found above, we can sketch the graph.

The Second Derivative, Concavity, and Points of Inflection

Up Next

Understanding Global Maximums and Minimums

Return to Module 2: Curve Sketching Using the Derivative