Question

Find the equation to the tangent and normal to the curve y = 3x^+x-2 at (1, 2).

Answer 1

Answer:

ok

Step-by-step explanation:

Answer 2

Answer:

Given curve is y=3x

2

−x+1

dx

dy

=6x−1, differentiate the curve

So slope of tangent at (1,3) is, m=

dx

dy

∣

∣

∣

∣

∣

(1,3)

=6(1)−1=5

Hence equation of tangent passes through (1,3) is given by

y−3=5(x−1)

⇒y−3=5x−5

⇒5x−y=2