Question

How do you find the equation of the line tangent to the graph of `f(t) = t − 13 t^2`, a = 3?

Answer 1

It is `y=-77x+87`

Explanation:

The general equation is

`y-f(a)=f'(a)(x-a)`

Hence we have that

`f(a)=f(3)=-114`

`f'(a)=1-26a=1-26*3=-77`

#y-f(a)=f'(a)(x-a)=>y-(-144)=(-77)(x-3)=>y+144=-77x+231=> y=-77x+87#