Question

How do you find the equation of a line tangent to the function `y=(2x+3)^2` at (-1,1)?

Answer 1

y = 4x +5

Explanation:

differentiate the function wrt x
we get 8x +12
now putting the value of x = -1
we get 4
that is the slope of the tangent at the point (-1,1)
using point slope form y-y1 = m(x-x1)
we get y-1 = 4(x+1)
giving y = 4x +5 which is the equation of the tangent