If
#color(white)("XXX")f(x)=1/x=x^(-1)#
then
#color(white)("XXX")f'(x)=(-1)(x^(-2)) = -1/x^2#
At #(color(red)(1/2),2)# the slope of the tangent line is
#color(white)("XXX")f'(color(red)(1/2))=-1/((color(red)(1/2))^2) = color(green)(-4)#
The slope-point form of the tangent line using
a slope of #color(green)(m=-4)#
and the point #(color(red)(barx),color(blue)(bary)) =(color(red)(1/2),color(blue)(2))#
is #color(white)("XXX")y-color(blue)(bary)=color(green)(m)(x-color(red)(barx))#
or, with the given values:
#color(white)("XXX")y-color(blue)(2=color(green)(-4)(x-color(red)(1/2))#
#rarr#
#color(white)("XXX")y-2 = -4x +2#
in standard form:
#color(white)("XXX")4x+y=4#
graph{(1/x-y)(4x+y-4)=0 [-3.973, 4.8, -1.268, 3.117]}
