Question 330494
The slope of the tangent line equals the value of the derivative at x=-1.
{{{y=x^3-5x+15}}}
{{{dy/dx=3x^2-5}}}
At {{{x=-1}}}
{{{y=(-1)^3-5(-1)+15=-1+5+15=-19}}}
{{{dy/dx=3(-1)^2-5=-2}}}
Use the point-slope form of a line, {{{y-yp=m(x-xp)}}}
{{{y-(-19)=-2(x-(-1))}}}
{{{y+19=-2(x+1)}}}
{{{y=-2x-2-19}}}
{{{y=-2x+17}}}
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That's what I get too. 
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{{{drawing(300,300,-5,5,-5,25,circle(-1,19,.2),grid(1),graph(300,300,-5,5,-5,25,x^3-5x+15,-2x+17))}}}