Problem 31 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{-3 x^{2}}{x^{2}+1}$$

Step-by-Step Solution

Answer

Domain: all real numbers; Symmetric about y-axis; Horizontal asymptote is \( y = -3 \).

Step 1: Identify the Domain

The function given is \( f(x) = \frac{-3x^2}{x^2 + 1} \). The denominator \( x^2 + 1 \) is never zero for any real number \( x \). Hence, the domain of \( f(x) \) is all real numbers, \( -\infty < x < \infty \).

Step 2: Determine x and y Intercepts

To find the x-intercepts, set \( f(x) = 0 \): \(-3x^2 = 0 \implies x = 0.\) However, \( -3x^2 \) cannot be zero for any non-zero \( x \), thus the x-intercept is at \( (0, 0) \). For the y-intercept, evaluate \( f(0)\): \(f(0) = \frac{-3(0)^2}{0^2 + 1} = 0.\) Thus, the y-intercept is also \( (0, 0) \).

Step 3: Analyze Symmetry

The function is considered even if \( f(-x) = f(x) \) for all \( x \). Calculating \( f(-x) = \frac{-3(-x)^2}{(-x)^2 + 1} = \frac{-3x^2}{x^2 + 1} \), we find that \( f(-x) = f(x) \). Therefore, the graph is symmetric with respect to the y-axis.

Step 4: Assess End Behavior

To understand end behavior, determine \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \). Calculating, \(\lim_{x \to \pm \infty} \frac{-3x^2}{x^2 + 1} = -3.\) Thus, the horizontal asymptote is \( y = -3 \).

Step 5: Plot Function Behavior and Asymptotes

Visualize the behavior of the function.- At \( x = 0 \), the value is \( 0 \).- As \( x \to \pm \infty \), the graph approaches the horizontal asymptote \( y = -3 \).- The graph is symmetric about the y-axis.Combine these insights to sketch the curve, noting that it dips down to \( 0 \) at \( x = 0 \) and levels out along \( y = -3 \) as \( x \) grows positively and negatively.

Key Concepts

Understanding the Domain of a Function

The domain of a function is the complete set of possible values of the independent variable, typically represented by "x." For the function represented by \( f(x) = \frac{-3x^2}{x^2 + 1} \), finding the domain involves checking when the denominator might be zero, since division by zero is undefined in mathematics.

In this case, the denominator is \( x^2 + 1 \), which is never zero for any real number \( x \) because \( x^2 \) is always non-negative, and adding 1 keeps it positive. This implies that \( f(x) \) is valid for all real numbers. Hence, the domain is all real numbers, expressed as \( (-\infty, \infty) \).

Understanding domain ensures that we only work with the values that give valid outputs, keeping our calculations mathematically sound.

Exploring Intercepts of a Graph

Intercepts are points where the graph meets the axes. To find the x-intercepts, set \( f(x) = 0 \). For \( f(x) = \frac{-3x^2}{x^2 + 1} \), setting the numerator \(-3x^2 = 0 \) implies \( x = 0 \); thus, the x-intercept is at the origin \((0, 0)\).

For the y-intercept, we substitute \( x = 0 \) into the function. This yields \( f(0) = \frac{-3 \times 0^2}{0^2 + 1} = 0 \), confirming the y-intercept is also at \((0, 0)\).

This double intercept at the origin tells us a lot about where the graph begins its journey across the coordinate plane.

Determining Symmetry of Functions

Symmetry in functions often makes them easier to analyze and draw. A graph is symmetric about the y-axis if the function is even, meaning \( f(-x) = f(x) \) holds for all \( x \).

Checking for the function \( f(x) = \frac{-3x^2}{x^2 + 1} \), compute \( f(-x) = \frac{-3(-x)^2}{(-x)^2 + 1} \), which simplifies back to \( f(x) \). Since \( f(-x) = f(x) \) is verified, the function is even.

This symmetry means the graph mirrors itself on either side of the y-axis, suggesting a balanced distribution of its shape.

Carrying Out End Behavior Analysis

End behavior describes how a function behaves as the value of \( x \) approaches positive or negative infinity. For \( f(x) = \frac{-3x^2}{x^2 + 1} \), we use limits to determine this behavior.

As \( x \to \pm \infty \), the dominant term in both the numerator and denominator is \( x^2 \). Hence, calculating \( \lim_{x \to \pm \infty} \frac{-3x^2}{x^2 + 1} \) involves simplifying to \( \lim_{x \to \pm \infty} \frac{-3x^2}{x^2(1 + \frac{1}{x^2})} = -3 \).

Thus, the horizontal asymptote is \( y = -3 \), which shows that as \( x \) becomes very large in the positive or negative direction, \( f(x) \) settles near \(-3\), outlining the graph's tendency at its extremes.

Understanding Asymptotes in Graphs

Asymptotes are lines that the graph approaches but never actually touches. For rational functions, horizontal asymptotes are significant, especially for end behavior.

In the function \( f(x) = \frac{-3x^2}{x^2 + 1} \), we've established that the horizontal asymptote is \( y = -3 \).

• Observe that as \( x \to \infty \) or \( x \to -\infty \), \( f(x) \) nears \(-3\).
• The graph levels off following closely along this line.

Understanding an asymptote's role helps us visualize the function's long-term trend and behavior.