Question
Find the equation of the tangent to the curve y = (x3 – 1) (x – 2) at points where the curve cuts the x-axis.
Solution
The equation of given curve is
It meets x-axis where y = 0
Putting y = 0, we get,
At (1, 0), slope of tangent = 4 (1)3 – 6 (1)2 –1 = 4 – 6 – 1 = – 3
At (2, 0), slope of tangent = 4 (2)3 – 6 (2)2 – 1 = 4 × 8 – 6 × 4 – 1 = 7
∴ equation of tangent at (1, 0) with slope – 3 is
y – 0 = – 3(x – 1) or y = – 3x + 3 or 3x + y – 3 = 0
The equation of tangent at (2, 0) with slope 7 is
y – 0 = 7(x – 2), or y = 7x – 14 or 7x – y - 14 = 0
