Question 1178952

{{{y=(3x^2-4x)/(2x+1)}}}  tangent at{{{ x=-1}}}


first find a slope and tangent point


{{{y}}}'{{{(x) = (6x^2 + 6 x - 4)/(2 x + 1)^2}}}

substitute {{{x=-1}}}

{{{y}}}'{{{(-1) =(6(-1)^2 + 6 (-1) - 4)/(2 (-1) + 1)^2}}}

{{{y}}}'{{{(-1) = -4}}}->Note this only tells us the slope of the line, not the equation for the line itself.


since the line is tangent to the function at {{{ x=-1}}}, by finding the {{{y}}}-coordinate we would possess a set of coordinates and a slope, allowing us to use point-slope form


{{{y=(3*(-1)^2-4*(-1))/(2*(-1)+1) =-7}}}

tangent point is ({{{-1}}},{{{-7}}}) and slope is{{{m= -4}}}


use point-slope form


{{{y-y[1]=m(x-x[1])}}}

{{{y-(-7)=-4(x-(-1))}}}

{{{y+7=-4(x+1)}}}

{{{y=-4x-4-7}}}

{{{y=-4x-11}}}



{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-1,-7,.12),locate(-1,-7,p(-1,-7)),

graph( 600, 600, -10, 10, -10, 10,(3x^2-4x)/(2x+1), -4x-11)) }}}