Question 315279
The slope of the tangent of the parabola is equal to the value of the derivative.
{{{y=3x^2+4x+1}}}
{{{dy/dx=6x+4=-2}}}
{{{6x+4=-2}}}
{{{6x=-6}}}
{{{x=-1}}}
So at the point {{{x=-1}}}, the slope of the tangent line equals -2.
At that point,
{{{y=3x^2+4x+1=3-4-1=-2}}}
So then
{{{y=-2x+b}}}
{{{-2=-2(-1)+b}}}
{{{b=-4}}}
and the equation of the tangent line is,
{{{y=-2x-4}}}
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{{{ drawing( 300, 300, -3, 3, -5, 5, grid(1),circle(-1,-2,.15),graph( 300, 300, -3, 3, -5, 5, 3x^2+4x-1, -2x-4)) }}}