Final answer:
To find the required tangent equation, we calculate the slope of the given line, find the derivative of f(x), set the derivative equal to the slope to solve for x, find the corresponding y-value on the curve, and then use the point-slope form to write the equation of the tangent line.
Step-by-step explanation:
To find the equation of the line that is tangent to the graph of f(x) = x3 and parallel to the line 12x - y + 3 = 0, we first need to determine the slope of the given line. The line can be rewritten in slope-intercept form as y = 12x + 3, which reveals a slope of 12. Since the tangent line must be parallel to this line, it will also have a slope of 12.
Next, we find the derivative of f(x) to determine where the slope of the tangent is also 12. The derivative f'(x) = 3x2 gives us the slope of the tangent line at any point on f(x). Setting f'(x) = 12 and solving for x gives us x = 2 (since 3x2 = 12 implies x2 = 4, so x = ±2 but we take x = 2 as it is the positive root).
Plugging x = 2 into f(x) to find the corresponding y value gives us f(2) = 23 = 8. Therefore, the point of tangency is (2, 8). The equation of the tangent line using the point-slope form is y - 8 = 12(x - 2), which simplifies to y = 12x - 16.