Find K in Tangent Line Equations: A Step-by-Step Guide

How to Find K in Tangent Line Equations

This guide will walk you through the process of finding the value of an unknown constant, ‘k’, within a function’s equation when you are given the equation of its tangent line at a specific point. We will use the relationship between the derivative of a function and the slope of its tangent line to solve for ‘k’.

Prerequisites

• Basic understanding of derivatives and differentiation rules.
• Familiarity with the slope-intercept form of a linear equation (y = mx + b).

Steps

1. Identify the Slope of the Tangent Line

   The problem provides the equation of the tangent line in slope-intercept form: y = mx + b. The slope of this line is represented by ‘m’. In this specific problem, the tangent line equation is given as y = 20x – 51. Therefore, the slope ‘m’ is 20.

2. Relate Slope to the Derivative

   A fundamental concept in calculus is that the slope of the tangent line to a function f(x) at a specific point x = a is equal to the value of the derivative of the function at that point, f'(a). The problem states that the tangent line touches the function f(x) at x = 3. This means that the slope of the tangent line, which we identified as 20, is equal to f'(3).

   So, we have: f'(3) = 20.

3. Find the Derivative of the Function f(x)

   The function is given as f(x) = kx^2 – kx – 15. To find f'(x), we need to apply the power rule and the constant multiple rule of differentiation.

   • The derivative of kx^2 is k * (2x^(2-1)) = 2kx.
   • The derivative of -kx is -k * (1x^(1-1)) = -k * 1 = -k.
   • The derivative of a constant (-15) is 0.

   Combining these, the derivative of f(x) is f'(x) = 2kx – k.

4. Evaluate the Derivative at the Given Point

   We know that f'(3) = 20. Now, substitute x = 3 into the derivative equation we found in the previous step:

   f'(3) = 2k(3) – k

   Simplify the expression: 2k(3) = 6k.

   So, f'(3) = 6k – k.

   Combine like terms: 6k – k = 5k.

   Therefore, f'(3) = 5k.

5. Solve for k

   Now we have two expressions for f'(3): f'(3) = 20 (from step 2) and f'(3) = 5k (from step 4). We can set these two expressions equal to each other to solve for k:

   5k = 20

   To isolate k, divide both sides of the equation by 5:

   k = 20 / 5

   k = 4

Conclusion

By understanding that the slope of the tangent line is equivalent to the derivative of the function at the point of tangency, we were able to set up an equation and solve for the unknown constant ‘k’. In this case, the value of k is 4.

Expert Note

Always double-check your differentiation steps. A small error in calculating the derivative can lead to an incorrect value for ‘k’. Ensure you are applying the power rule and constant multiple rule correctly.

Source: How to Find the Value of K Given the Equation of the Tangent Line (YouTube)