K well this is another one...yuck
The sketch shows the curve
y=x2−5and the normal at
(2,−6)a) Find the equation of the normal at the point
(2,−6)So I did
y=x^2 -5
dy/dx = 2x
2x=0
2(2)= 4 (The two coming from the given point)
so 4 is the gradient of the tangent (?)
Then -1/4 is the gradient of the normal (negative reciprocal)
So y= -1/4x + c
-6= -1/4 (2) + c
-11/2= c
Therefore
y= -1/2x + 11/2 for part a?
@Zacken