Finding The Value Of P For A Stationary Point On The X-axis
In calculus, stationary points, also known as critical points, are points on a curve where the derivative is equal to zero. These points are crucial in determining the local maxima, minima, and saddle points of a function. In this article, we will explore the process of finding the value of p for which the curve defined by the equation y = x³ + 3x² - 9x + p has a stationary point on the x-axis. This involves understanding the relationship between derivatives, stationary points, and the x-axis intercept.
Understanding Stationary Points
Stationary points, as the cornerstone concept, on a curve occur where the gradient of the curve is zero. Mathematically, this means that the first derivative of the function, denoted as dy/dx, is equal to zero. These points are significant because they indicate where the function momentarily stops increasing or decreasing. In simpler terms, a stationary point represents a turning point on the curve, potentially marking a local maximum, local minimum, or a saddle point.
To find these points, we first need to differentiate the equation of the curve with respect to x. For the given equation y = x³ + 3x² - 9x + p, the derivative, dy/dx, is found by applying the power rule of differentiation to each term. This rule states that if y = axⁿ, then dy/dx = naxⁿ⁻¹. Applying this to our equation, we get:
dy/dx = 3x² + 6x - 9
Now, to locate the stationary points, we set dy/dx to zero and solve for x:
3x² + 6x - 9 = 0
This is a quadratic equation, which can be simplified by dividing all terms by 3:
x² + 2x - 3 = 0
This quadratic equation can be factored into:
(x + 3)(x - 1) = 0
This gives us two solutions for x:
x = -3 and x = 1
These are the x-coordinates of the stationary points on the curve. To find the corresponding y-coordinates, we would substitute these values back into the original equation y = x³ + 3x² - 9x + p. However, in this specific problem, we are interested in the case where the stationary point lies on the x-axis, which means the y-coordinate must be zero. The value of p will be crucial in determining whether the curve intersects the x-axis at these stationary points.
The Condition for a Stationary Point on the x-axis
The x-axis is defined by the equation y = 0. Therefore, for a stationary point to lie on the x-axis, both the derivative dy/dx and the original function y must be equal to zero at that point. We have already found the x-coordinates of the stationary points by setting dy/dx to zero. Now, we need to ensure that the y-coordinate is also zero at these points. This means substituting the x-values we found (x = -3 and x = 1) into the original equation y = x³ + 3x² - 9x + p and setting the result to zero.
Let's start with x = -3:
0 = (-3)³ + 3(-3)² - 9(-3) + p 0 = -27 + 27 + 27 + p 0 = 27 + p p = -27
Now, let's check with x = 1:
0 = (1)³ + 3(1)² - 9(1) + p 0 = 1 + 3 - 9 + p 0 = -5 + p p = 5
We have found two possible values for p: -27 and 5. This means that the curve will have a stationary point on the x-axis when p is either -27 or 5. Each value corresponds to a different scenario where the curve touches the x-axis at a turning point. It's important to note that a cubic equation can have up to three real roots, and the stationary points can provide insight into the nature of these roots. If a stationary point lies on the x-axis, it indicates that the curve has a repeated root at that x-coordinate.
Determining the Value of p
To determine the specific value of p that satisfies the condition of a stationary point on the x-axis, we substitute the x-coordinates of the stationary points (x = -3 and x = 1) into the original equation y = x³ + 3x² - 9x + p and set y equal to zero. This is because a point on the x-axis has a y-coordinate of zero. We have already performed these calculations in the previous section, but let's reiterate the process for clarity.
For x = -3:
0 = (-3)³ + 3(-3)² - 9(-3) + p 0 = -27 + 27 + 27 + p p = -27
This tells us that when p = -27, the curve has a stationary point at x = -3 on the x-axis. This means the curve touches the x-axis at x = -3 and changes direction (either from increasing to decreasing or vice versa) at this point. This corresponds to a repeated root of the cubic equation.
For x = 1:
0 = (1)³ + 3(1)² - 9(1) + p 0 = 1 + 3 - 9 + p p = 5
This indicates that when p = 5, the curve has a stationary point at x = 1 on the x-axis. Similar to the previous case, the curve touches the x-axis at x = 1 and changes direction. This also corresponds to a repeated root of the cubic equation. Therefore, the values of p for which the curve has a stationary point on the x-axis are -27 and 5.
Conclusion
In conclusion, the problem demonstrates the application of calculus concepts, particularly differentiation and the identification of stationary points, to solve a geometric problem. We found that the values of p for which the curve y = x³ + 3x² - 9x + p has a stationary point on the x-axis are p = -27 and p = 5. This involved finding the derivative of the function, setting it to zero to find the x-coordinates of the stationary points, and then substituting these x-values into the original equation with y = 0 to solve for p. This method effectively combines algebraic manipulation with calculus techniques to arrive at the solution. Understanding these concepts is fundamental in analyzing the behavior of curves and functions in calculus.
