Question

What is the equation of the line tangent to `f(x)=1/sqrt(x^2+3x+6)` at `x=-1`?

Answer 1

`y= -1/16x + 7/16`

Explanation:

Start by finding the y-coordinate of intersection.

`f(-1) = 1/sqrt((-1)^2 + 3(-1) + 6)`

`f(-1) = 1/sqrt(4)`

`f(-1) = 1/2`

Now, differentiate.

`f(x) = 1/sqrt(x^2 + 3x + 6)`

`f(x) = 1/(x^2 + 3x + 6)^(1/2)`

`f(x) = (x^2 + 3x + 6)^(-1/2)`

Let `y = u^(-1/2)` and `u = x^2 + 3x+ 6`. Then `dy/(du) = -1/2u^(-3/2)` and `(du)/dx = 2x + 3`.

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx= -1/2u^(-3/2) xx (2x + 3)`

`dy/dx= -(2x + 3)/(2(x^2 + 3x + 6)^(3/2)`

The slope at `x= 1` is given by:

`dy/dx|_(x = -1) = -(2(-1) + 3)/(2(-1^2 + 3(-1) + 6)^(3/2))`

`dy/dx|_(x= -1) = -1/16`

The equation of the tangent line is therefore:

`y - y_1 = m(x- x_1)`

`y - 1/2 =-1/16(x - (-1))`

`y - 1/2 = -1/16x - 1/16`

`y = -1/16x + 7/16`

Hopefully this helps!