Question

How do you find the equation of the line tangent to `y = (5+6x)^2` at the point (-4,361) ?

Answer 1

Equation of tangent is `228x+y+551=0`

Explanation:

As slope of tangent to a curve at a given point is give by the value of first derivative at that point, let us first find the differential of `y=(5+6x)^2`.

As `y=(5+6x)^2`

`(dy)/(dx)=2(5+6x)xx6=60+72x` and at `(-4,361)` slope is `60-288=-228`

Hence equation of tangent is

`y-361=-228(x-(-4))`

or `228x+y-361+912=0`

or `228x+y+551=0`

graph{(228x+y+551)(y-(5+6x)^2)=0 [-10, 10, -20, 380]}