Given function:
#f(x)=4x^3-30x^2+48x#
The slope of tangent #dy/dx# at any point to the given curve is given by differentiating #f(x)# w.r.t. #x# as folows
#dy/dx=f'(x)#
#=d/dx(4x^3-30x^2+48x)#
#=12x^2-60x+48#
But the horizontal tangent has zero slope i.e. #dy/dx=0#
#therefore 12x^2-60x+48=0#
#x^2-5x+4=0#
#x^2-x-4x+4=0#
#x(x-1)-4(x-1)=0#
#(x-1)(x-4)=0#
#x=4, 1#
setting these values of #x# in given function we get corresponding values of y-coordinates as follows
#f(4)=4(4)^3-30(4)^2+48(4)=-32#
#f(1)=4(1)^3-30(1)^2+48(1)=22#
Hence, the coordinates of the points where tangent is horizontal, are #(4, -32)# & #(1, 22)#
Now, the equations of horizontal tangents with slope #m=0# at the points #(4, -32)# & #(1, 22)# are given by following formula
#y-y_1=m(x-x_1)#
#y-y_1=(0)(x-x_1)#
#y-y_1=0#
Setting #y_1=-32# & #y_1=22# in above equation we get equations of horizontal tangents as follows
#y-(-32)=0#
#y+32=0# &
#y-22=0#
