Final answer:
To find the equation of the tangent line to the curve r(x) = x³ - 12x² - 6x - 2 at the point where x = 0, we need to find the slope of the curve at that point. The slope of a curve at a specific point can be found by taking the derivative of the function and evaluating it at that point. Using the point-slope form of the equation of a line, we can find the equation of the tangent line.
Step-by-step explanation:
To find the equation of the tangent line to the curve r(x) = x³ - 12x² - 6x - 2 at the point where x = 0, we need to find the slope of the curve at that point. The slope of a curve at a specific point can be found by taking the derivative of the function and evaluating it at that point.
First, we take the derivative of the given function: r'(x) = 3x² - 24x - 6. Then, we evaluate the derivative at x = 0 to find the slope of the tangent line: r'(0) = 3(0)² - 24(0) - 6 = -6.
Now that we have the slope of the tangent line, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is the point on the line and m is the slope.
Since we want the tangent line at x = 0, the point on the line is (0, r(0)). We substitute this point and the slope into the point-slope form to get the equation of the tangent line:
y - r(0) = -6(x - 0)
Simplifying the equation gives us the final answer: y = -6x + r(0).