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How to Find the Slope and Equation of the Tangent Line for the Curve y 1 - 2x^3 at the Point (1, -1)

March 07, 2025Technology4335
How to Find the Slope and Equation of the Tangent Line for the Curve y

How to Find the Slope and Equation of the Tangent Line for the Curve y 1 - 2x3 at the Point (1, -1)

When working with calculus and understanding the behavior of functions, it's essential to be familiar with finding the slope of a curve and determining the equation of the tangent line at a specific point. In this article, we will explore how to find the slope of the curve y 1 - 2x3 at the point (1, -1), as well as the equation of the tangent line at the same point. This will involve using derivatives, the power rule, and the point-slope form of a linear equation.

The Derivative and Slope of the Curve

The slope of a curve at a specific point can be found using the derivative of the function. For the function y 1 - 2x3, we start by finding its derivative with respect to x.

Step 1: Finding the Derivative

The given function is y 1 - 2x3. We need to apply the power rule to find y'. The power rule states that if y axn, then y' naxn-1.

Applying the power rule to the function:

y'  0 - (2 × 3)x3-1    -6x2

Step 2: Evaluating the Derivative at x 1

Once we have the derivative y' -6x2, we can find the slope of the curve at x 1. Substitute x 1 into the derivative:

y'  -6(1)2    -6

Therefore, the slope of the curve at the point (1, -1) is -6.

Determining the Equation of the Tangent Line

Now that we have the slope of the tangent line at the point (1, -1), we need to find the equation of the tangent line using the point-slope form of a linear equation:

y - y1 m(x - x1)

Step 1: Identifying the Known Values

We know the slope m -6 and the point (1, -1). Substituting these values into the point-slope form:

y - (-1)  -6(x - 1)y   1  -6(x - 1)

Step 2: Simplifying to General Form

Next, we simplify the equation to the general form:

y   1  -6(x - 1)y   1  -6x   6y  -6x   5

Thus, the equation of the tangent line at the point (1, -1) is y -6x 5 (or 5 - 6x in standard form).

Conclusion

To summarize, the slope of the curve y 1 - 2x3 at the point (1, -1) is -6. The equation of the tangent line at this point is y -6x 5. These calculations involve calculus concepts such as derivatives and the point-slope form of a linear equation, which are fundamental in analyzing and understanding the behavior of functions.

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