find the equation of the normal to the graph of the function f(x)=-x^2+8x+4 at the point where x=-1

Question

find the equation of the normal to the graph of the function

Answer 1

Differentiate to find the slope at x=-1:

f'(x)=-2x+8 and f'(-1)=10.

The slope of the normal is therefore -1/10 and the equation of the normal is given by y=n(x)=-x/10+c, where c has to be found. f(-1)=1-8+4=-3. The normal must pass through the point (-1,-3) so put these values into the equation of the normal: -3=n(-1)=1/10+c, so c=-3.1 and n(x)=-(x/10+3.1). 3.1 can also be written 31/10 so n(x)=-(1/10)(x+31) or 10y+x+31=0.