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

Feb 7, 2016

$y = - \frac{3}{2} x + \frac{1}{2}$

#### Explanation:

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

$f \left(1\right) = 2 - 1 - \sqrt{2 \left(1\right) + 2} = 1 - \sqrt{4} = 1 - 2 = - 1$

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 $f ' \left(x\right)$ so we need to find the derivative of the function. Differentiating:

$f ' \left(x\right) = - 1 - 2 \cdot \frac{1}{2 \sqrt{2 x + 2}} = - 1 - \frac{1}{\sqrt{2 x + 2}}$

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

$f ' \left(1\right) = - 1 - \frac{1}{\sqrt{4}} = - 1 - \frac{1}{2} = - \frac{3}{2}$

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 $y = m x + c$ then solve for c:

$\left(- 1\right) = - \frac{3}{2} \left(1\right) + c \to c = \frac{3}{2} - 1 = \frac{1}{2}$

Thus we now have the equation of the tangent line:

$y = - \frac{3}{2} x + \frac{1}{2}$

The graph below will further illustrate the situation.