How do I find an equation of the tangent line to the curve at the point (2, 4) for the curve defined by y = 4x – 3x²?

To find the equation of the tangent line to the curve defined by the equation y = 4x – 3x² at the point (2, 4), we need to follow a few steps:

1. Differentiate the curve: We first calculate the derivative of the function, which gives us the slope of the tangent line. The derivative of the function y = 4x – 3x² is:
• y’ = 4 – 6x
2. Evaluate the derivative at x = 2: To find the slope of the tangent line at the point (2, 4), we substitute x = 2 into the derivative we calculated:
• y'(2) = 4 – 6(2) = 4 – 12 = -8
3. Use the point-slope form of the line: Now that we have the slope (m = -8) and the point (2, 4), we can use the point-slope form of the equation of a line, which is:
• y – y₁ = m(x – x₁)
4. Plug in the values: Substituting the point (2, 4) and the slope (-8) into the point-slope formula:
• y – 4 = -8(x – 2)
5. Simplify the equation: Distributing the -8 and solving for y:
• y – 4 = -8x + 16
• y = -8x + 20

The equation of the tangent line to the curve at the point (2, 4) is y = -8x + 20.