Help me asap please!

Question

find the equation of the tangent line to the curve y = f (x) at the given point. y=f(x)=4-x², at (3,-5)

Answer 1

f'(x) = -2x = slope of a tangent line at a point (x,y)

f'(3) = -2(3) = -6 = slope of the tangent line at the point (3,-5)

y+5 = -6(x-3) is the equation of the tangent line in point slope form

y= -6x +18-5

y =-6x +13 is the equation of the tangent line in slope intercept form

6x +y = 13 is the equation of the tangent line in standard form