Question

What is the equation of the line that is normal to `f(x)= (2-x)- sqrt( 2x+2)` at `x=1`?

Answer 1

`y = -3/2x+1/2`

Explanation:

Firstly we need a point at which the line intersects. We know that the line is tangent to `f(x)` at `x=1` so to find the corresponding `y` value simply substitute this into `f(x)` to obtain:

`f(1) = 2-1-sqrt(2(1)+2)=1-sqrt4=1-2=-1`

So we know the line passes through #(1,-1).

Next we will look at the gradient of the line. The gradient of the tangent is given by `f'(x)` so we need to find the derivative of the function. Differentiating:

`f'(x) = -1-2*1/(2sqrt(2x+2))=-1-1/sqrt(2x+2)`

We have used the chain rule to differentiate the second term. Now, substitute in a value of `x=1` and we get:

`f'(1) = -1-1/sqrt(4)=-1-1/2=-3/2`

Now that we have our gradient and a point the line passes through, the last thing to do is find the `y` inctercept. Substitute our values into `y=mx+c` then solve for c:

`(-1) = -3/2(1)+c -> c = 3/2-1=1/2`

Thus we now have the equation of the tangent line:

`y = -3/2x+1/2`

The graph below will further illustrate the situation.