What is the equation of the line normal to  f(x)=-sqrt((x-5)(x-2) at  x=6?

1 Answer
Dec 31, 2016

$y = \frac{4}{5} x - \frac{34}{5}$. See the normal-inclusive Socratic graph.

Explanation:

$f \le 0$.
At $x = 6 , f = - 2$. So, the point is $P \left(6 , - 2\right)$
To make f real,

$\left(x - 5\right) \left(x - 2\right) = {\left(x - \frac{7}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} \ge 0$, giving $x i \le 5 \mathmr{and} \le 2$

The Socratic graph is inserted, for yet another proof.

f'=-1/2(2x-7)/sqrt(((x-5)(x-2)(= -5/4#, at x = 6.

Slope of the normal is $- \frac{1}{f} ' \left(6\right) = \frac{4}{5}$.
\
So, the equation to the normal at $P \left(6 , - 2\right)$ is

$y - \left(- 2\right) = \frac{4}{5} \left(x - 6\right)$. Simplifying,

$y = \frac{4}{5} x - \frac{34}{5}$
graph{(y+sqrt((x-5)(x-2))) ( y-4/5x+34/5) = 0 [-10, 10, -5, 5]}