Question 1183487
given:  

{{{f(x) = -x^2 + 3x - 1}}}

Find: the equation of the tangent to the graph of {{{f(x) }}}at {{{x = 2}}}


{{{f(x) = -x^2 + 3x - 1}}}

{{{f}}}'{{{(x)=-2x + 3}}}

{{{f}}}'{{{(2)=-2*2 + 3=-1}}} => the slope of tangent line {{{m=-1}}}


{{{f(2) = -2^2 + 3*2 - 1=1}}}


=> tangent point is at ({{{2}}},{{{1}}})


tangent line: 

{{{y-y[1]=m(x-x[1])}}}-> plug in {{{m=-1}}} and ({{{2}}},{{{1}}})

{{{y-1=-1(x-2)}}}

{{{y-1=-1x+2}}}

{{{y= -x+3}}}



{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(2,1,.12), locate(2,-1,p(2,1)),

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