Derivative: Finding the normal line I have to find the normal line.
$$x^2-4x-5 ;(-2,7)$$
So, I have to find the equation of tangent line at $(x_1,y_1)$ which is,
$$2x_1-4$$
Now the equation at the (-2,7) is,
$$y-7=-8(x+2)$$
$$8x+y+9=0$$
Now, how do I find the normal line??
 A: Hint: The normal line is: $$y - 7 = m(x+2), m = -\dfrac{1}{2x_1-4} = \dfrac{1}{8}\implies 8y-56 = x + 2 \implies x - 8y +58 = 0$$
A: $$y=x^2-4x-5 \\\frac{dy}{dx}=\operatorname{slope}(x)=2x-4\\\text{ @ (-2, 7):}\\\frac{dy}{dx}\Big|_{x=-2,\;y=7}=-8=m\\\text{"normal" = perpendicular (}\perp\text{)}\\m_{\perp\xi}=-\frac1{m_\xi}\\\text{(slope of the line }\xi_\perp\text{, the line perpendicular to } \xi\text{ is the negative reciprocal of }\xi\text{'s slope)}\\\frac{dy_\perp}{dx}\Big|_{x=-2,\;y=7}=\frac18=m_\perp\\y_\perp=m_\perp x+b=\frac{x}8+b\\\text{If they meet at (-2,7):}\\7=\require{cancel}-\cancelto{\frac14}{\frac28}+b \\ \Rightarrow b=\frac{29}{4}$$
A: I'll betcha that the easiest way to do is to use the well-known fact that any normal to a line of slope $m$ has slope $-m^{-1}$.  This follows from the identity
$(\tan \theta)(\tan(\theta + \dfrac{\pi}{2})) = -1, \tag{1}$
which itself may be seen from
$\tan(\theta + \dfrac{\pi}{2}) = \dfrac{\sin(\theta + \frac{\pi}{2})}{\cos(\theta + \frac{\pi}{2})} = \dfrac{-\cos \theta}{\sin \theta} = -(\tan \theta)^{-1}; \tag{2}$
(2) follows from the angle addition formulas for $\sin$ and $\cos$, i.e. $\sin (\alpha + \beta) = (\sin \alpha)(\cos \beta) + (\sin \beta)(\cos \alpha)$ and so forth.  (Follows, that is, when $\theta \ne 0$.  For $\theta = 0$, take limits as $\theta \to 0$ or just create an exceptional case.)  In any event, (2) implies my previous assertion on slopes of perpendicular lines.  In the present case, the slope of the tangent is $-8$, so the slope of the normal is $1/8$.  Thus the equation of the normal line through $(-2,7)$ is
$y - 7 = \dfrac{1}{8}(x + 2) \tag{3}$
or
$y = \dfrac{1}{8} x + 7\frac{1}{4}. \tag{4}$
Now, any takers on my bet?
Hope this helps.  Good Solstice Season to All,
and as ever,
Fiat Lux!!!
