Answer:
(-2, -5), (2, 11)
Step-by-step explanation:
You want the coordinates of the points on the curve y = x³ +3 where the tangent lines are parallel to y = 12x -1.
Slope
The slope of the tangent line is the x-coefficient in its equation: 12.
The slope of the curve is given by its derivative:
y' = 3x²
We want the x-values where the slope is 12. These are the solutions to ...
12 = 3x²
4 = x² . . . . . . . . . . divide by 3
x = ±√4 = ±2 . . . . . take the square root
Coordinates
The coordinates of the points with x = ±2 are ...
y = (±2)³ +3 = ±8 +3 = {-5, 11}
The tangent points are (-2, -5) and (5, 11).
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Additional comment
The attached graphs and table show the solutions to y'=0 and the corresponding point locations. It also shows the tangent line equations (in point-slope form).
