Problem 11 - Calculus with Concepts in Calculus

Question

Sketch the graph of the function. $$ y=\frac{1-x^{2}}{x+1} $$

Step-by-Step Solution

Answer

The graph has a vertical asymptote at \( x = -1 \), an oblique asymptote \( y = -x \), crosses the y-intercept at (0, 1) and the x-intercept at (1, 0).

Step 1: Determine the Domain

The function is of the form \( y = \frac{1 - x^2}{x + 1} \). To find the domain, we need to identify where the function is undefined. The function is undefined where the denominator is zero, i.e., when \( x + 1 = 0 \). Solving for \( x \) gives \( x = -1 \). Therefore, the domain is all real numbers except \( x = -1 \).

Step 2: Analyze Asymptotes

Examining vertical asymptotes, we find a vertical asymptote at \( x = -1 \), where the denominator goes to zero. For horizontal asymptotes, as \( x \to \pm\infty \), the function resemblance becomes that of the leading terms: \( y \approx -x \). Hence, there is no horizontal asymptote, but we have an oblique/slant asymptote of \( y = -x \).

Step 3: Calculate Intercepts

For the y-intercept, set \( x = 0 \). \( y = \frac{1 - 0^2}{0 + 1} = 1 \). The y-intercept is \( (0, 1) \). For x-intercepts, set \( y = 0 \) which implies \( 1 - x^2 = 0 \), hence \( x = \pm 1 \). But since \( x = -1 \) is not in the domain, the only x-intercept is at \( (1, 0) \).

Step 4: Draw the Graph Considering Asymptotes and Intercepts

Consider the asymptotes, intercepts, and note the behavior of the function around \( x = -1 \). The function approaches negative and positive infinity as it nears this vertical asymptote from the left and right, respectively. Near \( x = 1 \), the curve crosses the x-axis. Sketch curves on both sides of the vertical asymptote while approaching the oblique asymptote of \( y = -x \).

Step 5: Verify Considerations and Complete Graph

Check that the curvature and intercepts are correct. Verify that the function approaches the oblique asymptote at \( y = -x \) at infinity, and adheres to the vertical asymptote at \( x = -1 \). The complete sketch reflects the intercepts, domain behavior, and asymptotic behavior described.

Key Concepts

Domain of a Function

The domain of a function refers to all the input values (usually represented as "x") for which the function is defined. For rational functions like the one we're working with, \( y = \frac{1 - x^2}{x + 1} \), the domain is determined by the denominator. Since division by zero is undefined, we set the denominator equal to zero and solve: \ \(x + 1 = 0\). Solving this, we find that \(x = -1\).
Therefore, the function is undefined at \(x = -1\), which means this value is excluded from the domain.
In simpler terms, the domain consists of all real numbers except where the denominator equals zero. For this function, that's all real numbers except \(x = -1\).
So the domain is: \(-\infty, -1) \cup (-1, \infty)\) instead of being all real numbers. Always ensure to recheck such calculation as this directly influences the function's behavior on the graph.

Vertical Asymptotes

Vertical asymptotes occur where a function tends toward infinity as the input approaches a particular value. We identify vertical asymptotes in rational functions by looking where the denominator is zero, given the simplest form of the function. For our function \(y = \frac{1 - x^2}{x + 1}\), we already calculated that the denominator becomes zero when \(x = -1\). This point of undefined behavior creates a vertical asymptote.
As you sketch the graph, you'll notice that as \(x\) gets closer to \(-1\), \(y\) will rapidly increase or decrease (approach infinity) without bound. In the graph, the curve will never touch or cross the line \(x = -1\).
Understanding vertical asymptotes is crucial because it helps in predicting the behavior of graphs and ensuring accurate representation in problem-solving.

X and Y Intercepts

Intercepts are points where the graph crosses the x-axis or y-axis. Finding these points help in sketching the graph properly.
**Y-Intercept:** To find the y-intercept, set \(x = 0\) in the equation. For \(y = \frac{1 - x^2}{x + 1}\), plugging \(x = 0\) gives \(y = 1\). Hence, the y-intercept is \((0, 1)\). This is where the graph meets the y-axis.
**X-Intercepts:** To find x-intercepts, set \(y = 0\) and solve for \(x\). The equation \(0 = 1 - x^2\) implies \(x^2 = 1\). Thus, \(x = \pm 1\). However, since \(x = -1\) is not in the domain, the only valid x-intercept for the graph is \( (1, 0) \). At this point, the graph crosses the x-axis.
Intercepts are pivotal in defining the starting points for drawing the curve for the function.

Oblique Asymptotes

Oblique or slant asymptotes are lines that a graph approaches as \(x\) tends toward infinity or negative infinity. They typically occur in rational functions where the degree of the numerator is higher than the denominator by one.For our function \(y = \frac{1 - x^2}{x + 1}\), as \(x \to \pm\infty\), the rational function behaves similarly to \(-x\). This indicates an oblique asymptote: \(y = -x\). To find this, divide the polynomial \(1-x^2\) by \(x+1\) using long division, which confirms \(y \approx -x\) for large \(|x|\). Unlike vertical asymptotes, the graph can approach and become arbitrarily close to an oblique asymptote.
This guidance line of \(y = -x\) helps in sketching the overall curve correctly, ensuring the function's end behavior is accurately portrayed, and captures the true path as \(x\) grows large positively or negatively.