Question 669056
{{{(f(x[2])-f(x[1]))/(x[2]-x[1])=(f(3)-f(0))/(3-0)}}}

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

{{{f(3)=-2*9+3 }}}

{{{f(3)=-18+3 }}}

{{{f(3)=-15 }}}

now find

{{{f(0)=-2*0^2+0 }}}

{{{f(0)=0 }}}


so,  the points are:

({{{0}}}, {{{f(0)}}}) =({{{0}}}, {{{0}}})

and ({{{3}}}, {{{f(3)}}}) =({{{3}}}, {{{-15}}})


now find a slope:

{{{f(3)-f(0)/(3-0)=(-15-0)/3=-15/3=-5}}}........the slope  is {{{m=-5}}}

Equation of the line thru ({{{0}}},{{{0}}}) and with slope {{{m=-5}}}

Equation Form: {{{y = mx + b}}}

{{{0 = -5*0 + b}}}

{{{b = 0}}}

Equation of the secant: {{{y = -5x}}}

here is a graph:

{{{ graph( 600, 600, -10, 10, -20, 10,-2x^2+x,-5x) }}}