Question

Find an equation of the tangent line to the curve at the point (0, 1).

y = (1 + 3x)20

i got y' = 60(3x+1)^19

what do i do next?

Answer 1

at the point (0,1) x=0, so sub that in

y' = 60(1)^19 = 60

but (0,1) is a y-intercept of 1

so equation is y = 60x + 1

Answer 2

y' = 60(3x + 1)^19
y' = 60(3(0) + 1)^19
y' = 60(1)=60

Then use point-slope form
y - y1 = 60(x - x1)
y - 1 = 60(x - 0)
y - 1 = 60x
y = 60x + 1