Question

Find the equation of the tangent line to the function f(x) =x³_2x²_3x at the point (1, _4) give your answer in the form y=mx+b

Answer 1

To find the equation of the tangent line to the function f(x) = x³ - 2x² - 3x at the point (1, -4), we need to find the derivative, evaluate it at x = 1, and use the point-slope form of a linear equation.

Step-by-step explanation:

To find the equation of the tangent line to the function f(x) = x³ - 2x² - 3x at the point (1, -4), we need to find the derivative of the function and evaluate it at x = 1. The derivative of f(x) is f'(x) = 3x² - 4x - 3. Evaluating f'(x) at x = 1, we get f'(1) = 3(1)² - 4(1) - 3 = -4. Therefore, the slope of the tangent line is -4.

Now, we can use the point-slope form of a linear equation y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope. Substituting the values, we have y - (-4) = -4(x - 1). Simplifying, we get y + 4 = -4x + 4. Rearranging the equation, we have y = -4x.

Therefore, the equation of the tangent line to the function f(x) = x³ - 2x² - 3x at the point (1, -4) is y = -4x.