Question

What is the equation of the tangent line of `f(x)=(2x+1)(x+2)` at `x=2`?

Answer 1

`y= 13x - 6`

Explanation:

First, simplify the function through distribution so we can differentiate it easier.

`f(x)=2x^2+5x+2`

We should find the point of tangency:

`f(2)=2(4)+5(2)+2=20`

The tangent line will pass through the point `(2,20)`.

Through the power rule, we know that

`f'(x)=4x+5`

The slope of the tangent line will be equal to the value of the derivative at `x=2`, or

`f'(2)=4(2)+5=13`

We know the tangent line has a slope of `13` and passes through the point `(2,20)`.

We can write this as an equation in `y=mx+b` form. So far, we know that `m=13`.

`y=13x+b`

Substitute in `(2,20)` for `(x,y)`.

`20=13(2)+b`

`b=-6`

Thus, the equation of the tangent line is

`y=13x-6`

Graphed are `f(x)` and its tangent line:

graph{((2x+1)(x+2)-y)(y-13x+6)=0 [-4, 6, -10, 50]}