Let y=-x^(2) + 6, find the equation of the tangent line that passes through the point (4,0). Hint: the tangent line has the equation y-0=m(x-4)

Question

I don't know where to start.

Answer 1

The derivative will give you the slope of the tangent line at that point.

 

y = -x^2 + 6

 

y' = -2x

 

At x=4,  the slope is M = f'(4) =  (-2)(4) = -8

 

The tangent line has slope M=-8 and passes through (4,0)

 

y - 0 = (-8)(x-4)  is the point slope form as specified.

 

y = (-8)(x-4)

 

y = -8x + 32