Problem 54
Question
Show that the curve \(y=\sqrt{x^{2}+4 x}\) has two slant asymptotes: \(y=x+2\) and \(y=-x-2 .\) Use this fact to help sketch the curve.
Step-by-Step Solution
Verified Answer
The curve \(y=\sqrt{x^2+4x}\) has slant asymptotes \(y=x+2\) and \(y=-x-2\).
1Step 1: Understand the Curve Equation
The curve is given as \(y = \sqrt{x^2 + 4x}\). To find slant asymptotes, we should analyze the behavior of the function as \(x\) approaches infinity or negative infinity.
2Step 2: Factor Inside the Square Root
Let's rewrite \(x^2 + 4x\) as \((x+2)^2 - 4\). This decomposition will help in simplifying expressions later. This turns the function into \(y = \sqrt{(x+2)^2 - 4}\).
3Step 3: Analyze Behavior as \(x \to \infty\)
For large \(x\), the expression \((x+2)^2 - 4\) simplifies approximately to \((x+2)^2\). Thus, \(y \approx \sqrt{(x+2)^2} = |x+2|\). For large positive \(x\), \(|x+2| = x+2\). Hence, one asymptote is \(y = x + 2\).
4Step 4: Analyze Behavior as \(x \to -\infty\)
Similarly, for large negative \(x\), \(|x+2| = -(x+2)\). So \(y \approx -x-2\). Therefore, the other asymptote is \(y = -x - 2\).
5Step 5: Sketch the Curve
The curve approaches the line \(y = x + 2\) as \(x \to \infty\) and the line \(y = -x - 2\) as \(x \to -\infty\). To sketch the curve, note that it will lie above \(y = 0\) for all real \(x\), and it will transition smoothly between the two asymptotes as \(x\) ranges from \(-\infty\) to \(\infty\).
Key Concepts
Asymptotic BehaviorCurve SketchingLimit Analysis
Asymptotic Behavior
Asymptotic behavior gives us a glimpse into how a function behaves as the input becomes very large or very small. In this case, the function is given by \( y = \sqrt{x^2 + 4x} \). Identifying asymptotes helps understand the general direction in which the curve heads.
- A slant asymptote typically occurs when, for a rational function, the degree of the polynomial in the numerator is one more than the degree of the polynomial in the denominator.
- For the given equation, directionality as \( x \to \infty \) helps discover the slant asymptotes: \( y = x + 2 \) as \( x \to \infty \) and \( y = -x - 2 \) as \( x \to -\infty \).
- This indicates that as \( x \) becomes very large (positively or negatively), the curve will closely follow these linear paths.
Curve Sketching
When sketching curves, understanding their asymptotic behavior is fundamental. It helps define the visual boundary and overall direction. The curve described by \( y = \sqrt{x^2 + 4x} \) serves as a valuable case.
- First, identify the vertical and horizontal tends. This is crucial as they assist in predicting behaviors across the plane.
- Since the function includes a square root, the results must be non-negative. Therefore, the graph will remain above the \( y = 0 \) line.
- The transition between slant asymptotes \( y = x + 2 \) and \( y = -x - 2 \) requires a smooth path. The curve will arc smoothly, starting above the more negative asymptote and changing direction as it heads toward the more positive one.
Limit Analysis
Limit analysis is the mathematical approach to understanding the behavior of a function as the input approaches a certain value. In this context, we're interested in the values as \( x \to \infty \) or \( x \to -\infty \).
- By expressing \( x^2 + 4x \) as \((x+2)^2 - 4\), and looking at the limit as \( x \to \infty \), the equation simplifies to \( y \approx x + 2 \).
- Similarly, analyzing as \( x \to -\infty \) results in \( y \approx -x - 2 \). This simplification helps identify asymptotic trends.
- Such limits are instrumental in pinpointing where the function is heading and gauging the function's end behavior.
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