What is the equation of the line that is normal to f(x)= sqrt( x^2+3x+2)  at  x=1 ?

Feb 8, 2018

$y = \frac{\sqrt{6}}{5} \left(- 2 x + 7\right)$
See explanation below

Explanation:

The value of derivative of $f \left(x\right)$ at $x = 1$ is the slope of tangent line to $f \left(x\right)$ at $x = 1$.

As per, $f ' \left(x\right) = \frac{1}{2} \frac{2 x + 3}{\sqrt{{x}^{2} + 3 x + 2}}$

f'(1)=1/2(2·1+3)/sqrt(1^2+3·1+2)=$\frac{5}{2 \sqrt{6}}$

By other hand: if the straigh line general equation is

$y = m x + b$ (where $m$ is the slope and $b$ is the intersection with $y$ axis) then, the normal line equation is $y = - \frac{1}{m} x + c$

So: the normal line equation requested is $y = - 2 \frac{\sqrt{6}}{5} x + b$

The value b can be found by the fact: the value of $f \left(1\right)$ and by normal line must be the same (because they intercept in $x = 1$

Thus: $f \left(1\right) = \sqrt{6} = - 2 \frac{\sqrt{6}}{5} + b$ trasposing elements we found

$b = 7 \frac{\sqrt{6}}{5}$

Finally, the normal line equation is $y = - 2 \frac{\sqrt{6}}{5} x + 7 \frac{\sqrt{6}}{5}$ or

$y = \frac{\sqrt{6}}{5} \left(- 2 x + 7\right)$