Question 1155282
 Find the slope of the line tangent to the following at point, where {{{x=2}}}

{{{f(x)=e^(-0.1x^2+2x-3)}}}.......if {{{x=2}}}

{{{f(x)=e^(-0.1*2^2+2*2-3)}}}

{{{f(x)=e^(-0.1*4+4-3)}}}

{{{f(x)=e^(-0.4+1)}}}

{{{f(x)=e^(0.6)}}}

so, the point is ({{{2}}},{{{e^0.6}}}) or ({{{2}}},{{{1.8}}}) 


compute slope {{{m}}}

take first derivative

{{{df(x)/dx=e^(-0.1x^2+2x-3)*(-0.2x+2)}}}

if {{{x=2}}}

{{{m=e^(-0.1*2^2+2*2-3)*(-0.2*2+2)}}}

{{{m=2.91539}}}


{{{y=mx+b}}}

{{{y=2.91539x+b}}}........use point ({{{2}}},{{{e^0.6}}})to find {{{b}}}

{{{e^0.6=2.91539*2+b}}}

{{{b=e^0.6-5.83078}}}

{{{b=1.8221188003905089-5.83078}}}

{{{b=- 4.00866}}}

and, tangent is:

{{{y = 2.91539x - 4.00866}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,1.82,.12),locate(2,1.82,p(2,1.8)),
 graph( 600, 600, -10, 10, -10, 10, e^(-0.1x^2+2x-3), 2.91539x - 4.00866)) }}}