Question 417649
{{{f(x)=-3(x-2)^2+1}}}

find 2 points 


{{{f(x)=-3(x^2-4x+2)+1}}}


{{{f(x)=-3x^2+12x-6+1}}}

{{{f(x)=-3x^2+12x-5}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .....{{{a=-3}}}}, {{{b=12}}} and {{{c=-5}}}


{{{x = (-12 +- sqrt( 12^2-4*(-3)*(-5) ))/2*(-3) }}} 


{{{x = (-12 +- sqrt( 144 -60 ))/(-6) }}}



{{{x = (-12 +- sqrt( 84 ))/(-6) }}}



{{{x1 = (-12 +- 9.2) /-6 }}}



{{{x1 = (-2.8) /-6 }}}



{{{x1 = 0.47}}}...one {{{x-intercept}}}....point:(0.47,0)


{{{x2 = (-12 - 9.2) /-6 }}}



{{{x2 = (-21.2) /-6 }}}



{{{x2 = 3.53}}}...the other {{{x-intercept}}}....point:(3.53,0)




{{{ graph( 500, 500, -10, 10, -10, 10, -3x^2+12x-5) }}}