Question

What is the equation of the line tangent to `f(x)=(x+6)^2-7x` at `x=-1`?

Answer 1

Tangent Line `y=3x+35`

Explanation:

Start from the given function

`f(x)=(x+6)^2-7x` at `x=-1`

find the ordinate first using x=-1

`f(-1)=(-1+6)^2-7(-1)`

`f(-1)=32`

Let `(x_1, y_1)=(-1, 32)`

slope for the slope `m` by obtaining the first derivative of `f(x)`

`f(x)=(x+6)^2-7x`
`f' (x) =2(x+6)^(2-1)-7`

slope `m=f' (-1)=2(-1+6)-7=10-7=3`

`m=3`

Solve for the tangent line

`y-y_1=m(x-x_1`

`y-32=3(x- -1)`

`y=3x+3+32`

`y=3x+35" "`the tangent line

kindly see the graph of `f(x)=(x+6)^2-7x` and the tangent line `y=3x+35`

God bless....I hope the explanation is useful.