Question

determine the following characteristics of the graph.
y = \frac{x^2}{x^2 + 3}
type dne if the answer does not exist for any of the characteristics below

• domain
• intercepts
• x - intercept(s)

if there are multiple points, separate with a comma. ex, (-9,0),(4,0)

• y - intercept
• symmetry

select an answer

• asymptotes
• horizontal asymptote
• vertical asymptote
• intervals of increase and decrease
• interval of increase
• interval of decrease
• local max and/or min
• list all maximum points
• list all minimum points

Explanation:

Step1: Find Domain

The function is \( y = \frac{x^2}{x^2 + 3} \). The denominator \( x^2 + 3 \) is always positive (since \( x^2 \geq 0 \), so \( x^2 + 3 \geq 3 > 0 \) for all real \( x \)). Thus, the domain is all real numbers.
Expression: Domain: \( (-\infty, \infty) \)

Step2: Find x - intercepts

Set \( y = 0 \), so \( \frac{x^2}{x^2 + 3} = 0 \). This implies \( x^2 = 0 \) (since denominator can't be zero), so \( x = 0 \). The x - intercept is \( (0, 0) \).
Expression: \( x \) - intercept: \( (0, 0) \)

Step3: Find y - intercept

Set \( x = 0 \) in the function: \( y=\frac{0^2}{0^2 + 3}=0 \). So the y - intercept is \( (0, 0) \).
Expression: \( y \) - intercept: \( (0, 0) \)

Step4: Check Symmetry

Replace \( x \) with \( -x \): \( y=\frac{(-x)^2}{(-x)^2 + 3}=\frac{x^2}{x^2 + 3} \), which is the same as the original function. So it is symmetric about the y - axis.
Expression: Symmetry: Symmetric about y - axis

Step5: Find Horizontal Asymptote

For rational functions, compare degrees of numerator and denominator. Both are degree 2. The horizontal asymptote is \( y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{1}{1} = 1 \).
Expression: Horizontal Asymptote: \( y = 1 \)

Step6: Find Vertical Asymptote

Denominator \( x^2 + 3 \) never equals zero for real \( x \), so no vertical asymptote (DNE).
Expression: Vertical Asymptote: DNE

Step7: Find Intervals of Increase/Decrease

First, find the derivative. \( y=\frac{x^2}{x^2 + 3} \), using quotient rule \( y'=\frac{2x(x^2 + 3)-x^2(2x)}{(x^2 + 3)^2}=\frac{6x}{(x^2 + 3)^2} \).
Set \( y' = 0 \), we get \( 6x = 0\Rightarrow x = 0 \).
Test intervals:

• For \( x < 0 \) (e.g., \( x=-1 \)), \( y'=\frac{-6}{(1 + 3)^2}<0 \), so function is decreasing on \( (-\infty, 0) \).
• For \( x > 0 \) (e.g., \( x = 1 \)), \( y'=\frac{6}{(1 + 3)^2}>0 \), so function is increasing on \( (0, \infty) \).

Expression: Interval of Increase: \( (0, \infty) \); Interval of Decrease: \( (-\infty, 0) \)

Step8: Find Local Max/Min

From the derivative, at \( x = 0 \), the function changes from decreasing to increasing, so it is a local minimum. The local minimum point is \( (0, 0) \), no local maximum.
Expression: Local Minimum: \( (0, 0) \); Local Maximum: DNE

Answer:

• Domain: \( (-\infty, \infty) \)
• \( x \) - intercept(s): \( (0, 0) \)
• \( y \) - intercept: \( (0, 0) \)
• Symmetry: Symmetric about y - axis
• Horizontal Asymptote: \( y = 1 \)
• Vertical Asymptote: DNE
• Interval of Increase: \( (0, \infty) \)
• Interval of Decrease: \( (-\infty, 0) \)
• Local maximum points: DNE
• Local minimum points: \( (0, 0) \)