Question 1009964
The slope in general would be {{{df/dx=m=((x^2+x+1)(3x^2)-(x^3+1)(2x+1))/(x^2+x+1)^2}}}

the steps to simplify not shown here,...


{{{m=(x^4+2x^3+3x^2-2x-1)/(x^2+x+1)^2}}}.


For the point whose x coordinate is 1,
{{{f(1)=y=(1^3+1)/(1^2+1+1)}}}
{{{f(1)=y=2/4=1/2}}}
the point being  (1, 1/2).


The slope for x=1, according to the formula found for {{{m=df/dx}}}, 
is  {{{(1^3+2*1^3+3*1^2-2*1-1)/(2^2)}}}
{{{3/4}}},


Then if using point-slope linear equation form,
{{{highlight(y-1/2=(3/4)(x-1))}}}
or put into slope-intercept form,
{{{y=(3/4)x-3/4+1/2}}}
{{{y=(3/4)x-3/4+2/4}}}
{{{highlight(y=(3/4)x-1/4)}}}


The other question can then be handled much the same, just re-using the formula for the slope and the given point.




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The mnemonic for formula of derivative of a rational function goes as  {{{(ho*dhi-hi*dho)/(ho*ho)}}} .