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

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

1. Find the derivative: The derivative of a function gives us the slope of the tangent line at any point on that curve. Let’s find the derivative of the function:
• Given: y = 4x – 3x²
• Using the power rule, the derivative is:
• dy/dx = 4 – 6x

Now we can find the slope of the tangent line at the point (2, 4):

1. Evaluate the derivative at x = 2:
• Substituting x = 2 into the derivative:
• dy/dx = 4 – 6(2) = 4 – 12 = -8

So, the slope of the tangent line at the point (2, 4) is -8.

1. Use the point-slope form to write the equation of the tangent line: The point-slope form of the line is given by:

y – y₁ = m(x – x₁)

Where:

• (x₁, y₁) is the point (2, 4) and m is the slope, which we found to be -8.

Plugging in these values:

y – 4 = -8(x – 2)

Now, let’s simplify this equation:

1. Multiply both sides by -1:

y – 4 = -8x + 16

Adding 4 to both sides, we get:

y = -8x + 20

The equation of the tangent line to the curve y = 4x – 3x² at the point (2, 4) is:

y = -8x + 20