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Curve Sketching as its name suggests helps us sketch the approximate graph of any given function which can further help us visualize the shape and behavior of a function graphically. Curve sketching isn’t any sure-shot algorithm that after application spits out the graph of any desired function but it is an active role approach for a visual representation of a function that needs analysis of various features of graphs, such as intercepts, asymptotes, extrema, and concavity, to gain a better understanding of how the function behaves.

In this article, we will explore all the fundamentals of curve sketching and its solved examples. Other than that we will also explore all the aspects in detail which will help us analyze and sketch the function more efficiently.

Curve Sketching Definition

Curve Sketching is a collection of various techniques which can be used to create the approximate graph of any given function. That can help us analyze different features and behavior of the graph. Curve Sketching involves analysis of many aspects of a given function such as changes in function as input changes, maximum and minimum values, intercepts, domain, range, asymptotes, etc. Curve Sketching is used to visualize and understand the shape and behavior of any given function.

Basic of Curve Sketching

 

Graphing Basics

To create a graph of any given function, we need to plot some points such as intercepts, critical points, and some regular points which can help us trace the graph on the cartesian plane. Let’s further understand these basics in detail as follows:

Plotting Points

We can easily plot various different points of any function on the graph by just using random input and their outputs as the coordinates. This random plot of points helps us connect the final graph after all the necessary calculations are done. For example, we need to graph the function f(x) = ex, so just putting x = loge3 we get the output f(loge3) = 3. Now, we can (loge3, 3) as a point on the graph. 

Domain and Range

First, analyze the function to check for its domain. We need to find out the points where the value of the function becomes undefined or is discontinuous. For example: 

1/x is not defined at x = 0. Log(x) is defined only at positive values of x. 

Finding Intercepts and Asymptotes

Intercepts are the points where the graph cuts the coordinate axis and to find the x-intercept, we put y = 0 and solve for x. Similarly, to find the y-intercept, we put x = 0 and solve for y.

Asymptotes are lines that the graph approaches but do not intersect. There are three types of asymptotes which are as follows: 

  • Horizontal Asymptote
  • Vertical Asymptote
  • Slant Asymptote

Horizontal asymptotes occur when the function approaches a constant value as x approaches infinity or negative infinity. Vertical asymptotes occur when the function approaches infinity or negative infinity as x approaches a specific value. Slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator.

To calculate Horizontal Asymptote, we need to calculate the limit of a function at infinity, and vertical asymptotes are those points for which functions become not defined i.e., the denominator becomes 0.

Learn more about, Vertical and Horizontal Asymptotes

Example: Find Intercept and Asymptote for f(x) = (2x + 1) / (x – 3).

Solution:

To find the x-intercept, we set f(x) = 0 and solve for x:

⇒ (2x + 1) / (x – 3) = 0

⇒ 2x + 1 = 0 (x ≠ 3)

⇒ x = -1/2

Therefore, the x-intercept of the function is at (-1/2, 0).

To find the y-intercept, we set x = 0 and solve for f(x):

⇒ f(0) = (2(0) + 1) / (0 – 3) = -1/3

Therefore, the y-intercept of the function is at (0, -1/3).

The vertical asymptote occurs at x = 3, since the denominator of the function becomes zero at that point.

To find the horizontal asymptote, we need to examine the behavior of the function as x approaches infinity or negative infinity. We can do this by dividing the numerator and denominator by the highest power of x in the function:

f(x) = (2x + 1) / (x – 3) = (2 + 1/x) / (1 – 3/x)

As x becomes very large or very small, the term 1/x becomes insignificant compared to the other terms in the numerator and denominator, so we can ignore it:

f(x) ≈ 2 / 1 = 2 (as x → ±)

Therefore, the horizontal asymptote of the function is y = 2.

Local Extrema and Inflection Points

Local Extrema are those points of the function or graph for which there is no such value of function greater or smaller than the local extrema i.e., no other point in the neighborhood of the local extrema has a more extreme value than it. 

To find out the maxima and minima in any function, we need to find the critical points. Critical points of the function are defined as the points where either slope of the function is not defined or the slope is 0 i.e., f'(x) = 0.

After getting the values of critical points, check the second derivative of the function at those critical points. If f”(x) > 0 for some critical point x=k, then f(k) is the local minima of the function, and if f”(x) < 0 for some critical point x = k, then f(k) is the local maxima of the function. 

If f”(x) = 0 for some critical point x = k then x = k is the Point of Inflection or Inflection Point of the function.

Calculating Slope and Concavity

The slope is the measure of inclination from the positive x-axis and it tells us whether the graph is increasing (slope>0) or decreasing (slope<0). To find the slope of any given function, we differentiate the given function w.r.t to the dependent variable and substitute the value for which we need to calculate the slope.

Concavity is the measure of the curve which tells us whether the graph is concave up or concave down i.e., the direction of curvature of the graph. To calculate the concavity, we use the second derivative w.r.t dependent variable of the function. The second derivative tells us the rate at which the derivative is changing. If the second derivative is positive, then the function is concave up, and if the second derivative is negative, then the function is concave down.

Example: Find the slope and concavity of f(x) = x3 – 3x2 + 2x.

Solution:

f(x) = x3 – 3x2 + 2x

⇒ f'(x) = 3x2 – 6x + 2

To find the slope of the function at a specific point, we substitute the value of x in the derivative:

f'(-1) = 3(-1)^2 – 6(-1) + 2 = 11

Therefore, the slope of the function at x = -1 is 11.

To find the concavity of the function,

f”(x) = 6x – 6

To find the points where the concavity changes, we set f”(x) = 0 and solve for x:

⇒ f”(x) = 6x – 6 = 0

⇒ x = 1

Therefore, the function changes from concave down to concave up at x = 1.

Here, (1, 0) is the inflection point.

Inflection Point: Inflection point is the point where the concavity changes i.e., the second derivative of function = 0. 

Symmetry 

Determine whether the functions are odd, even, or neither of these. Sometimes some functions are periodic in nature. We need to check for their periodicity if they are periodic in nature. Functions satisfying, f(x) = f(-x) are called even functions. While the functions satisfying f(-x) = -f(x) are called odd functions. Some examples of periodic functions are: 

Sin(x), Cos(x), and other trigonometric functions. 

Tracing Different Types of Functions

We can sketch any function using Curve Sketching but some functions can be quite tricky to sketch. Let’s consider some examples of various different functions which we will sketch using the techniques of curve sketching.

Linear Functions

Sketching Linear Function is quite an easy task in curve sketching as we just need two points on the graph and the line joining those two points is the graph of a linear function. Let’s consider an example:

Example: Sketch the graph for the function, f(x) = 2x + 3.

Solution:

For f(x) = 2x + 3,

Put x = 0 ⇒ f(0) = 3

and Put x = 1 ⇒ f(1) = 5

Now, draw a striaght line passing throught the pionts (0, 3) and (1, 5) which is the graph of the linear function f(x) = 2x+3.

Graph for f(x) = 2x + 3

 

Polynomial Functions

Polynomial functions occur a lot in calculus, and it is essential to know how to sketch their graphs. We will look at a function and use the techniques studied above to infer the graph of the function. The general idea is to look for asymptomatic values, and where they are going, and then find the critical points and draw a graph according to them. Let’s see it through examples, 

Example: Sketch the graph for the given function, 

f(x) = x2 + 4

Solution: 

We know that the domain of this function is all real numbers. This functions will tend to infinity as we go towards large positive and negative values of x. 

Notice that f(-x) = (-x)2 + 4 = x2 + 4 = f(x). That is this function is even, so its graph must be symmetric about the y-axis. 

Now we know that graph goes to infinity and is symmetrical around the y-axis. Now, let’s look for critical points. 

f'(x) = 2x = 0 

⇒ x = 0 

Thus, there is only one critical point which is x = 0. Checking the double derivative f”(x) = 2. Since f”(x) > 0 for every x. So, the graph must be convex upward everywhere with minima at x = 0. Now we just need to know the value of the function at minima. 

f(0) = 4. 

Now we are ready to plot a graph. 

Graph of f(x) = x^2 + 4

 

Exponential Functions

Exponential functions are an essential part of calculus and are commonly represented as f(x) = ax, where a is any positive constant and x can be any possible real number. To sketch the graph of Exponential Functions we need to check, the domain, range, and asymptotes. We also need to check whether the function is increasing or decreasing. If the base of the exponential function lies between 0 and 1 then it decreases in its domain otherwise it is an increasing function.

Let’s consider an example of sketching the exponential function.

Example: Sketch the graph for the given function, f(x) = 2x – 1

Solution:

The domain of this function is all real numbers. As x goes to negative infinity, the function approaches zero, and as x goes to infinity, the function approaches infinity.

The base of the exponential function is 2, which is greater than one, so the function increases as x increases.

To find critical points, we need to find where the derivative is zero. f'(x) = 2^x ln(2). This derivative is zero only when x = 0.

Thus, the critical point occurs at x = 0. To determine the concavity, we can find the second derivative of the function. f”(x) = 2^x ln^2(2). Since the second derivative is always positive, the graph must be convex upward everywhere.

Now we just need to find the value of the function at the critical point. f(0) = 2^0 – 1 = 0.

We can now use this information to sketch the graph of the exponential function.

Graph for f(x) = 2^x - 1

 

Logarithmic Functions

We know that logarithmic functions are inverse of exponential functions. The function y = logbx is the inverse of y = bx. The graph of the exponential function is given below. We also know that the graph of an inverse of a function is basically a mirror image of the graph in y = x. So we can derive the shape of the graph of log function from the given graph of the exponential function. 

The mirror image of the Logarithmic function is the exponential function both of them are shown in the image below,

Logarithmic Functions and Exponential Function

 

Let’s see an example of graphic logarithmic functions. 

Example: Plot the graph for log10x + 5. 

Solution: 

We can see that the function is f(x) = log10x + 5.

The graph of this equation will be shifted 5 units in the upwards direction.

Graph for log10x + 5.

 

Read More,

Sample Problems on Curve Sketching

Problem 1: Sketch the graph for the given function, 

f(x) = x + 8

Solution: 

We know that the domain of this function is all real numbers. This functions will tend to infinity as we go towards large positive and negative values of x. 

Now we know that graph goes to positive infinity for larger positive values of x and negative infinity for larger negative values of x.Now, let’s look for critical points. 

f'(x) = 1  

There is no critical point, that means derivatives change sign remains same and constant throughout. 

Let’s see where the equation cuts the x-axis. 

x+ 8 = 0 

⇒x = -8  

Now we are ready to plot a graph. 

Graph for f(x) = x + 8

 

Problem 2: Sketch the graph for the given function, 

f(x) = x2 – 6x + 8

Solution: 

We know that the domain of this function is all real numbers. This functions will tend to infinity as we go towards large positive and negative values of x. 

Now we know that graph goes to positive infinity for larger positive values of x and negative infinity for larger negative values of x.Now, let’s look for critical points. 

f'(x) = 2x -6 = 0 

⇒x = 3

There is one critical point, that means derivatives change sign at that, but we don’t know which sign changes to what. So, we will check the sign.

From x ∈ (-∞,3] f'(x) < 0. That is in this interval, the graph is decreasing. 

From x ∈ (3,∞) f'(x) > 0. That is in this interval, the graph is increasing.  

That means the critical point is a minimum. 

Let’s see where the equation cuts the x-axis. 

x2 -6x + 8 = 0 

⇒x2 -4x -2x + 8 = 0 

⇒x(x – 4) -2(x – 4) = 0

⇒(x – 2)(x – 4) = 0

Now we are ready to plot a graph. 

Graph for f(x) = x^2 - 6x + 8

 

Problem 3: Sketch the graph for the given function, 

f(x) = x3 – 3x + 4

Solution: 

We know that the domain of this function is all real numbers. This functions will tend to infinity as we go towards large positive and negative values of x. 

Now we know that graph goes to positive infinity for larger positive values of x and negative infinity for larger negative values of x.Now, let’s look for critical points. 

f'(x) = 3x2 -3 = 0 

⇒x2 = 1 

⇒x = -1 or 1

There are two critical points, that means derivatives change sign at them, but we don’t know which sign changes to what. So, we will check the sign.

From x ∈ (-∞,-1] f'(x) > 0. That is in this interval, the graph is increasing. 

From x ∈ (-1,1] f'(x) < 0. That is in this interval, the graph is decreasing.

From x ∈ (1,∞) f'(x) > 0. That is in this interval, the graph is increasing.  

f(0) = 4. 

Now we are ready to plot a graph. 

Graph for f(x) = x3 - 3x + 4

 

Problem 4: Plot the graph for the equation f(x) = e + 2. 

Solution: 

We know that f(x) = ex + 2 is an exponential function, it increases with increasing value of x. 

f'(x) = e

This will never become zero, so there are no critical points. The graph is continuously increasing. 

f”(x) > 0 thus it’s shape is always convex upward. Due to the addition of 2 to the exponential function. The whole graph will be shifted two units upwards. 

Graph for f(x) = e^x  + 2.

 

FAQs on Curve Sketching

Q1: Define Curve Sketching.

Answer:

Curve Sketching is the set of techniques used to approximate sketch of graph of any given function.

Q2: Why is Curve Sketching Important?

Answer:

Curve Sketching is important as it helps us visualize and understand properties and behavior of any function. By sketching the graph of any function, we can find out about a lot things such as maxima and minima value of the function, increasing or decreasing behavior etc.

Q3: What are Some Common Features of a Graph that can be Determined through Curve Sketching?

Answer:

Some common features of a graph that can be determined through curve sketching include the function’s

  • Intercepts
  • Extrema
  • Asymptotes
  • End Behavior

Q4: How do you Find the Intercepts of a Graph?

Answer:

To find the x-intercepts of a graph, you need to set the function equal to zero and solve for x and similarly to find the y-intercepts, you need to evaluate the value of y when x=0.

Q5: How do you Find the Extrema of a Graph?

Answer:

To find the extrema of a graph, you need to find the critical points of the function, which are the points where the derivative is equal to zero or does not exist. Then, you can use the first or second derivative test to determine whether these critical points correspond to local maxima, local minima, or neither.

Q6: What are Asymptotes?

Answer:

Asymptotes are imaginary lines that the graph of a function approaches but never touches. They can be vertical, horizontal, or oblique (slanted). Vertical asymptotes occur when the function approaches positive or negative infinity as x approaches a certain value. Horizontal asymptotes occur when the function approaches a constant value as x approaches positive or negative infinity. Oblique asymptotes occur when the function approaches a slanted line as x approaches positive or negative infinity.

Q7: How do you Determine the End Behavior of a Graph?

Answer:

To determine the end behavior of a graph, you need to examine what happens to the function as x approaches positive or negative infinity. If the function approaches a horizontal asymptote, its end behavior will be similar to the behavior of the asymptote. If the function approaches positive or negative infinity, its end behavior will be increasing or decreasing, respectively.



Last Updated : 12 Jun, 2023
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