Final answer:
To find equations for two lines that are both tangent to the curve y = x³ - 3x² + 3x − 8 and parallel to the line 3x - y = 12, we can use the derivative of the curve to find the points where the tangent lines intersect the curve. Finding two points and using the point-slope form of the equation of a line, we can determine the equations for the two tangent lines to be y = 3x + 9 and y = 3x + 7.
Step-by-step explanation:
To find equations for two lines that are both tangent to the curve y = x³ - 3x² + 3x − 8 and parallel to the line 3x - y = 12, we can use the fact that parallel lines have the same slope. The slope of the given line is 3, so we need to find two lines with a slope of 3 that are tangent to the curve. To do this, we can find the derivative of the curve and use the derivative to find the points where the tangent lines intersect the curve.
The derivative of y = x³ - 3x² + 3x − 8 is y' = 3x² - 6x + 3. Setting y' equal to 3 gives us the equation 3x² - 6x + 3 = 3. Solving this equation gives us x = 1. Substituting x = 1 into the original curve equation gives us y = 1³ - 3(1)² + 3(1) − 8 = -6. So the first tangent point is (1, -6).
Using this point and the slope of 3, we can find the equation of the first tangent line. The point-slope form of a line is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. Substituting the values, we have y - (-6) = 3(x - 1), which simplifies to y = 3x + 9.
Using the same process, we can find the second tangent point to be (2, -5), and the equation for the second tangent line is y = 3x + 7. Therefore, the equations for the two lines that are both tangent to the curve y = x³ - 3x² + 3x − 8 and parallel to the line 3x - y = 12 are y = 3x + 9 and y = 3x + 7.