Recall the point slope form of a line through(x0,y0)
y = m(x-x0) + y0
For the tangent line m = f '(x0) for the normal line m =-1/f ' (x0)
For the vertex don't forget your college algebra --complete the square!!
for the function f(x) = 2x^2 - 8x - 5
how do i find equation of tangent, and equation of normal line to the curve y = f(x) at (1, f(1)) ? and vertex?
as far as i know, for the normal line,
f(#), so it would be (1,f(#))
and - 1/f'(x) is normal line
but i don't know how to find "equation" of normal line, and equation of tangent. and vertex........?