Since the gradient of a line is a tangent ratio, we need to find first the tangent ratio of #135^@#
i.e. #tan(135^@)#
We find the derivative of #f(x)=x^2-x#. This allows us to find the gradient at any point on #f(x)=x^2-x#, and we know our result needs to be #tan(135^@)#
#dy/dx(x^2-x)=2x-1#
#:.#
#2x-1=tan(135^@)#
#x=(tan(135^@)+1)/2=(-1+1)/2=0#
For #60^@#
#2x-1=tan(60^@)#
#x=(tan(60^@)+1)/2=(sqrt(3)+1)/2#
Check:
Equation of tangent line for #bb(b)#
Plugging #0# into the derivative:
#2(0)-1=-1#
This is our gradient:
Corresponding #y# value:
#(0)^2-0=0#
Equation is:
#y-0=-(x-(0))#
#color(blue)(y=-x)#
For #bbc#
#2((sqrt(3)+1)/2)-1=sqrt(3)#
#y# value:
#((sqrt(3)+1)/2)^2-((sqrt(3)+1)/2)=1/2#
Equation:
#y-1/2=sqrt(3)(x-((sqrt(3)+1)/2))#
#color(blue)(y=sqrt(3)-1-(sqrt(3))/2)#
GRAPH:
