X-Coordinate On Curve Tangent Parallel To X-Axis
Hey guys! Let's dive into a cool math problem where we need to find the x-coordinate (also known as the abscissa) of a point on a curve. But not just any point – we're looking for the point where the tangent line to the curve is perfectly horizontal, meaning it's parallel to the x-axis. We’ll be working with the curve defined by the equation y = x³ + 4x² - 11x - 30, and we're interested in the interval [-2, 3]. Sounds like fun, right? Let's break it down step by step.
Understanding the Problem
So, what does it mean for a tangent line to be parallel to the x-axis? Think about it: a line parallel to the x-axis has a slope of zero. The slope of a tangent line at any point on a curve is given by the derivative of the curve's equation at that point. Therefore, our mission is to find the x-values within the interval [-2, 3] where the derivative of our curve y = x³ + 4x² - 11x - 30 equals zero. This is a classic calculus problem that combines differentiation with finding roots of an equation. We're essentially looking for the points where the curve momentarily stops increasing or decreasing and has a horizontal tangent. These points are critical in understanding the behavior of the function, especially in determining local maxima and minima. The interval [-2, 3] restricts our search, making sure we only focus on the relevant portion of the curve. It’s like saying, “Okay, we know the tangent is horizontal somewhere, but let's find it within this specific window.” This makes the problem more manageable and practical, as in real-world applications, we often care about the behavior of a function within certain limits. Finding these points helps us visualize the curve and its key characteristics, making it a valuable skill in various fields, from physics to economics. We'll use the power of calculus to zoom in on these special locations where the curve's slope flattens out, giving us a crucial insight into the function's landscape. So, let's get our calculus tools ready and embark on this mathematical adventure!
Step 1: Find the Derivative
Alright, first things first, we need to find the derivative of our function. Remember, the derivative, often denoted as dy/dx or y', gives us the slope of the tangent line at any point on the curve. For the function y = x³ + 4x² - 11x - 30, we'll use the power rule for differentiation, which states that the derivative of xⁿ is nx^(n-1). Applying this rule to each term in our function, we get:
- The derivative of x³ is 3x²
- The derivative of 4x² is 8x
- The derivative of -11x is -11
- The derivative of the constant -30 is 0
So, putting it all together, the derivative dy/dx is 3x² + 8x - 11. This new equation represents the slope of the tangent line at any given x-value on the original curve. Think of it as a slope-finding machine! You plug in an x-value, and it spits out the slope of the tangent at that point. This is a crucial step because now we have a way to mathematically describe the slope, which is the key to finding where the tangent is parallel to the x-axis. Understanding derivatives is fundamental in calculus, acting as a cornerstone for many applications in physics, engineering, and economics. By finding the derivative, we've transformed our original problem of finding a tangent line into a problem of solving an equation. We've essentially translated the geometric idea of a horizontal tangent into an algebraic expression, making it easier to handle. The derivative is not just a mathematical formula; it’s a powerful tool that allows us to analyze how a function changes, and in this case, it helps us identify those special points where the change is momentarily zero.
Step 2: Set the Derivative to Zero
Now that we have the derivative, dy/dx = 3x² + 8x - 11, we need to find the x-values where this derivative equals zero. Why? Because, as we discussed earlier, a tangent line parallel to the x-axis has a slope of zero. So, we're essentially finding the x-coordinates where the slope of the curve is zero, indicating a horizontal tangent. This translates to solving the quadratic equation 3x² + 8x - 11 = 0. There are several ways to solve a quadratic equation: factoring, using the quadratic formula, or completing the square. In this case, let's try factoring first, as it can be a quicker method if it works. Factoring involves breaking down the quadratic expression into two binomial expressions. We look for two numbers that multiply to give the product of the leading coefficient (3) and the constant term (-11), which is -33, and add up to the middle coefficient (8). These numbers are 11 and -3. So, we can rewrite the middle term 8x as 11x - 3x and then factor by grouping:
- 3x² + 11x - 3x - 11 = 0
- x(3x + 11) - 1(3x + 11) = 0
- (x - 1)(3x + 11) = 0
Setting each factor equal to zero gives us the solutions for x:
- x - 1 = 0 => x = 1
- 3x + 11 = 0 => x = -11/3
So, we have two potential x-coordinates where the tangent line might be parallel to the x-axis: x = 1 and x = -11/3. Solving this quadratic equation is a critical step because it directly provides the potential locations where our curve has a horizontal tangent. It’s like using a detective's magnifying glass to pinpoint the exact spots on the curve that meet our criteria.
Step 3: Check the Interval
Okay, we've found two potential x-values: x = 1 and x = -11/3. But remember, we're only interested in the interval [-2, 3]. This means we need to check if both of our solutions fall within this interval. x = 1 clearly falls within the interval [-2, 3], since -2 ≤ 1 ≤ 3. However, x = -11/3 is approximately -3.67, which is outside our interval because -3.67 < -2. So, we can disregard x = -11/3. This step is crucial because it ensures we're only considering solutions that are relevant to the specific part of the curve we're analyzing. It’s like putting on blinders to block out the extraneous information and focus only on what matters. In real-world applications, such constraints are common, reflecting the limitations or specific conditions of a problem. Checking the interval helps us avoid including solutions that, while mathematically correct, don't make sense in the context of our problem. This is a key aspect of problem-solving – not just finding the answers, but also ensuring they are meaningful and applicable. By checking the interval, we're refining our search, narrowing down the possibilities, and making sure our final answer is both accurate and relevant to the original question.
The Answer
Therefore, the x-coordinate (abscissa) of the point on the curve y = x³ + 4x² - 11x - 30 where the tangent is parallel to the x-axis within the interval [-2, 3] is x = 1. We've successfully navigated the problem, using the derivative to find the slope, setting it to zero to identify potential points, and checking the interval to ensure our solution is relevant. This journey highlights the power of calculus in understanding the behavior of functions and solving geometric problems. Remember, math isn't just about formulas and calculations; it's about understanding concepts and applying them logically. We took a geometric concept – a tangent line parallel to the x-axis – and translated it into an algebraic equation that we could solve. This is the essence of mathematical problem-solving: bridging the gap between concepts and calculations to arrive at a meaningful answer. So, pat yourselves on the back, mathletes! You've tackled this problem like pros, and you're one step closer to mastering the fascinating world of calculus.