Question

Find the equation of the tangent and normal to the curve:

$y = x^2 - 4x - 5[tex] at [tex]x = - 2$

Answer 1

y = x² - 4x - 5 put  x= -2
y = 4 + 8 - 5 = 7
we find at point (-2,7)
dy/dx = 2x - 4
dy/dx = -4 - 4 = -8                let m1                     (at point (-2,7))
equation of tangent 
y - 7 = -8(x + 2)
8x + y + 9 = 0           at point (-2,7)
slope of normal 
m1*m2 = -1
m2 = 1/8
equation of normal 
y - 7 = (x + 2)/8
x - 8y + 58 = 0