Curve sketching basics

You must know how to properly sketch a curve (or graph) of a function. This means identifying and drawing its roots, critical points, and asymptotes. You also need to correctly draw the curve's shape, by using what you know from the function's first and second derivatives to draw a correctly oriented slope and proper concavity.

In this chapter, we will provide you with a detailed step-by-step Curve Sketching Algorithm to easily draw any graph!

Curve sketching algorithm

1. Determine discontinuities and their behaviour

This includes point, infinite and jump discontinuities

For asymptotes, test their behaviours using limits

2. Determine critical points

Set the first derivative to zero and find roots and points that are undefined

3. Determine intervals of increase/decrease/neither

Calculate the sign of the first derivative on each interval

4. Test critical points for maximum/minimum/neither

5. Determine points of inflection

6. Determine intervals of concave up/down/neither

Calculate the sign of the second derivative on each interval

7. Sketch the graph

example

Graph the function $f(x) = \frac {x^2 + 2x - 3} {x^2 - 4}$ using the curve sketching algorithm

answer

This video will show you how to properly sketch a graph, just so that you can get a feel for how it is done:

Example

Use the curve sketching algorithm to sketch $f(x) = \frac {2(x^2 - 9)} {x^2 - 4}$.

Answer

To sketch the graph properly, we need to know its first and second derivatives:

$f'(x) = \frac {20x} {(x^2 - 4)^2}$

$f''(x) = \frac {-20(3x^2 + 4)} {(x^2 - 4)^3}$

1) Firstly, we need to determine the discontinuities for $f(x)$:

$f(x)$ is discontinuous at $x = -2$ and $x = 2$ since it has asymptotes there.

Let's determine the behaviours of the function at these asymptotes:

$^{\lim}_{x \to -2^-} (\frac {2(x + 3)(x - 3)} {(x - 2)(x + 2)}) = \frac {2(1)(-5)} {(-4)(0^-)} = - \infty \quad$ (note that $-2^- + 2 = 0^-$)

$^{\lim}_{x \to -2^+} (\frac {2(x + 3)(x - 3)} {(x - 2)(x + 2)}) = \frac {2(1)(-5)} {(-4)(0^+)} = + \infty \quad$ (note that $-2^+ + 2 = 0^+$)

$^{\lim}_{x \to 2^-} (\frac {2(x + 3)(x - 3)} {(x - 2)(x + 2)}) = \frac {2(5)(-1)} {(0^+)(4)} = + \infty \quad$ (note that $2^- - 2 = 0^-$)

$^{\lim}_{x \to 2^+} (\frac {2(x + 3)(x - 3)} {(x - 2)(x + 2)}) = \frac {2(5)(-1)} {(0^-)(4)} = - \infty \quad$ (note that $2^+ - 2 = 0^+$)

We should also take a look at the end behaviours for any horizontal asymptotes:

$^{\lim}_{x \to + \infty} \frac {2(x^2 - 9)} {x^2 - 4} = \ ^{\lim}_{x \to + \infty} \frac {2x^2 - 18} {x^2 - 4} = \ ^{\lim}_{x \to + \infty} \frac {2 - \frac {18} {x^2} } {1 - \frac {4} {x^2} } = \frac {2 - 0^+} {1 - 0^+s} = 2 \$ from below

$^{\lim}_{x \to - \infty} \frac {2(x^2 - 9)} {x^2 - 4} = \ ^{\lim}_{x \to + \infty} \frac {2x^2 - 18} {x^2 - 4} = \ ^{\lim}_{x \to + \infty} \frac {2 - \frac {18} {x^2} } {1 - \frac {4} {x^2} } = \frac {2 - 0^+} {1 - 0^+s} = 2 \$ from below

2) Now we need to determine the critical points of $f(x)$:

$f'(x)$ has one root at $x = 0$, and is also undefined at the asymptotes.

Then, $f(0) = \frac {2(0 - 9)} {0 - 4} = \frac {-18} {-4} = 4.5$.

So, $(0, 4.5)$ is a critical point, as well as $x = -2$ and $x = 2$.

3) Now we must find the intervals of increase/decrease/neither:

So, f(x) is decreasing when $x < 0$ and increasing when $x > 0$.

4) Based on the calculations we already made, the critical point$(9, 4.5)$ is a minimum since it decreases then increases.

5) Using the second derivative, we will now determine if there are any points of inflection:

$f''(x) = \frac {-20(3x^2 + 4)} {(x^2 - 4)^3}$

$3x^2 + 4 = 0 \Rightarrow x^2 \ne - \frac {4} {3} \Rightarrow f(x)$ does not have any POIs

6) Using the second derivative, we need to determine the concavity for every interval of $f(x)$:

So, $f(x)$ is concave down when $x < -2$ and $x > 2$, but concave up when $-2 < x < 0$ and $0 < x < 2$.

7) Finally, use all the information to properly sketch the graph: