Answer :
m at (x = 1) = 2
y at (x = 1) = (1)4 – 6(1)3 + 13(1)2 – 10(1) + 5 = 3
Tangent : y – b = m(x – a)
y – 3 = 2(x – 1)
2x – y + 1 = 0
Normal :
x + 2y – 7 = 0
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