If f'(-3)=2 and f(-3)=5, write the equation of the normal line to y=f(x) at x=-3

Answer 1

If f'(-3) = 2, then the slope of the tangent line at x = -3 is 2

The slope of the normal line = -1 / (the slope of the tangent line) = - 0.5

If f(-3) = 5, we can write the equation of the normal line to y=f(x) at x= - 3 (and y = 5) in the point-slope form as

y - 5 = -0.5 (x + 3) or in the slope-intercept form as y = - 0.5 x + 3.5