Graphing Y = 8x / (x^2 + 4) A Step-by-Step Guide

by THE IDEN 49 views

Introduction

In this comprehensive guide, we will delve into the process of graphing the function y = 8x / (x^2 + 4). This exploration involves a step-by-step analysis, including identifying the domain and any symmetries, calculating the first and second derivatives (y’ and y’’), pinpointing the critical points, and understanding the function's behavior at each critical point. By meticulously dissecting these aspects, we will construct an accurate and informative graph of the given function. This detailed approach will not only help in visualizing the function but also provide a deeper understanding of its mathematical properties and behavior. Understanding how to graph functions like this is a fundamental skill in calculus and is essential for various applications in science and engineering. Therefore, let’s embark on this journey to master the art of graphing rational functions, starting with the very basics and gradually building up to more complex concepts.

1. Domain and Symmetries

Determining the Domain

First and foremost, we need to determine the domain of the function. The domain is the set of all possible input values (x-values) for which the function is defined. For the given function, y = 8x / (x^2 + 4), we observe that the denominator is x^2 + 4. A rational function is undefined when the denominator is equal to zero. Therefore, we need to find the values of x for which x^2 + 4 = 0. Solving this equation, we get x^2 = -4. Since there are no real solutions for this equation (as the square of a real number cannot be negative), the denominator is never zero for any real value of x. Consequently, the domain of the function is all real numbers, which can be expressed in interval notation as (-∞, ∞). Understanding the domain is crucial because it tells us the range of x-values over which the graph will exist. It prevents us from looking for the function's behavior in regions where it is not defined.

Identifying Symmetries

Next, we investigate the symmetries of the function. Symmetries can simplify the graphing process by allowing us to predict the behavior of the function on one side of the y-axis based on its behavior on the other side. There are two primary types of symmetry to consider: even symmetry (symmetry about the y-axis) and odd symmetry (symmetry about the origin). To test for even symmetry, we replace x with -x in the function and see if we obtain the original function. So, let's evaluate y(-x):

y(-x) = 8(-x) / ((-x)^2 + 4) = -8x / (x^2 + 4) = -y(x)

Since y(-x) = -y(x), the function exhibits odd symmetry. This means that the function is symmetric about the origin. In other words, if (x, y) is a point on the graph, then (-x, -y) is also a point on the graph. This property will be invaluable when sketching the graph, as we only need to analyze the function's behavior for x ≥ 0 and then reflect it about the origin to obtain the complete graph. Recognizing symmetry not only makes the graphing process easier but also provides insights into the fundamental nature of the function. Odd symmetry, in particular, tells us that the function will pass through the origin, which is an important piece of information.

2. First Derivative (y’) and Critical Points

Calculating the First Derivative

The first derivative, denoted as y’, is a fundamental tool in calculus for understanding the rate of change of a function. It helps us determine where the function is increasing or decreasing and identify local maxima and minima. To find the first derivative of y = 8x / (x^2 + 4), we will use the quotient rule. The quotient rule states that if y = u(x) / v(x), then y’ = (u’(x)v(x) - u(x)v’(x)) / (v(x))^2. In our case, u(x) = 8x and v(x) = x^2 + 4. Let's find the derivatives of u(x) and v(x):

  • u’(x) = 8
  • v’(x) = 2x

Now, applying the quotient rule, we get:

y’ = (8(x^2 + 4) - 8x(2x)) / (x^2 + 4)^2

Simplifying the expression:

y’ = (8x^2 + 32 - 16x^2) / (x^2 + 4)^2

y’ = (-8x^2 + 32) / (x^2 + 4)^2

This is the first derivative of the function. The first derivative is a crucial element in understanding the function's behavior, as it provides information about the slope of the tangent line at any point on the graph. A positive first derivative indicates that the function is increasing, while a negative first derivative indicates that the function is decreasing. The points where the first derivative is zero or undefined are known as critical points, which we will explore next.

Identifying Critical Points

Critical points are the points where the first derivative is either equal to zero or undefined. These points are crucial because they are potential locations of local maxima, local minima, or points of inflection. To find the critical points, we set the first derivative y’ equal to zero and solve for x:

(-8x^2 + 32) / (x^2 + 4)^2 = 0

A fraction is equal to zero if and only if its numerator is equal to zero. Therefore, we set the numerator equal to zero:

-8x^2 + 32 = 0

Divide by -8:

x^2 - 4 = 0

Factor the quadratic:

(x - 2)(x + 2) = 0

This gives us two critical points:

  • x = 2
  • x = -2

Since the denominator (x^2 + 4)^2 is never zero for any real value of x, there are no additional critical points from the derivative being undefined. The critical points x = 2 and x = -2 are essential markers on the x-axis where the function's behavior might change. To fully understand what happens at these points, we need to evaluate the function y at these x-values and also analyze the sign of the first derivative in the intervals determined by these critical points. This will reveal whether these points correspond to local maxima, local minima, or neither.

3. Function's Behavior at Critical Points

Evaluating the Function at Critical Points

To understand the function's behavior at the critical points, we need to evaluate the function y = 8x / (x^2 + 4) at x = 2 and x = -2. This will give us the y-coordinates of these critical points, allowing us to plot them on the graph.

For x = 2:

y(2) = (8 * 2) / (2^2 + 4) = 16 / (4 + 4) = 16 / 8 = 2

So, the point (2, 2) is a critical point.

For x = -2:

y(-2) = (8 * -2) / ((-2)^2 + 4) = -16 / (4 + 4) = -16 / 8 = -2

Thus, the point (-2, -2) is another critical point.

These two points, (2, 2) and (-2, -2), are significant landmarks on the graph of the function. They are potential locations for local maxima or local minima. To determine whether these points are maxima, minima, or points of inflection, we need to analyze the sign of the first derivative around these points. This will tell us whether the function is increasing or decreasing on either side of these critical points.

Analyzing the First Derivative

To analyze the function's behavior around the critical points, we examine the sign of the first derivative, y’ = (-8x^2 + 32) / (x^2 + 4)^2, in the intervals determined by the critical points x = -2 and x = 2. The critical points divide the real number line into three intervals: (-∞, -2), (-2, 2), and (2, ∞). We will choose a test point within each interval and evaluate y’ at that point to determine its sign.

  1. Interval (-∞, -2): Choose a test point, say x = -3.

    y’(-3) = (-8(-3)^2 + 32) / ((-3)^2 + 4)^2 = (-8(9) + 32) / (13)^2 = (-72 + 32) / 169 = -40 / 169

    Since y’(-3) < 0, the function is decreasing in the interval (-∞, -2).

  2. Interval (-2, 2): Choose a test point, say x = 0.

    y’(0) = (-8(0)^2 + 32) / (0^2 + 4)^2 = (32) / (16) = 2

    Since y’(0) > 0, the function is increasing in the interval (-2, 2).

  3. Interval (2, ∞): Choose a test point, say x = 3.

    y’(3) = (-8(3)^2 + 32) / (3^2 + 4)^2 = (-8(9) + 32) / (13)^2 = (-72 + 32) / 169 = -40 / 169

    Since y’(3) < 0, the function is decreasing in the interval (2, ∞).

From this analysis, we can conclude the following:

  • At x = -2, the function changes from decreasing to increasing. Therefore, (-2, -2) is a local minimum.
  • At x = 2, the function changes from increasing to decreasing. Therefore, (2, 2) is a local maximum.

This information is crucial for sketching the graph, as it gives us the turning points of the function. We now know where the function reaches its highest and lowest points locally. To further refine our understanding of the graph's shape, we will examine the second derivative, which will provide insights into the concavity of the function.

4. Second Derivative (y’’) and Concavity

Calculating the Second Derivative

The second derivative, denoted as y’’, provides valuable information about the concavity of the function. Concavity refers to the direction in which the curve is bending. If the second derivative is positive, the function is concave up (shaped like a