Calculus 3 Question Please Explain

Find unit vector a) parallel to and b) perpendicular to f(x) at the given point. f(x) = -x^2 + 5. Point (1,4)
Thank you

First start by finding the slope of the tangent line at (1,4), so...

f '(x) = -2x
f '(1) = -2

The -2 is really -2/1 (change in y/change in x) so your parallel vector would be <1,-2>. Now you need to divide by the magnitude of the vector. ||<1,-2>|| = sqrt(5)

so your answer for part a) is <1,-2> / sqrt(5)

for part b) since you're in 2-d, you can just switch the x and y components of your vector and make either or x or y negative, so it can be either <-2,-1> / sqrt(5) or <2,1> / sqrt(5).

Hope that made sense ! :)

f(x) = -x^2 + 5
f'(x)=-2x
f'(1)=-2
a) slope of parallel vector -2
b) slope of perp. vector 1/2

a) unit vector (1i-2j)/sqrt(5)
b) unit vector (2i+j)/sqrt(5)