Question 651774
you have the quadratic in the form {{{ax^2+b+c}}}

set it equal to zero and solve for {{{x}}}

{{{-x^2+4x+1=0}}}...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ...note that {{{a=-1}}}, {{{b=4}}} and {{{c=1}}}

{{{x = (-4 +- sqrt( 4^2-4*(-1)*1 ))/(2*(-1)) }}}


{{{x = (-4 +- sqrt( 16+4 ))/(-2) }}}

{{{x = (-4 +- sqrt( 20))/(-2) }}}

{{{x = (-4 +-  4.47 )/(-2) }}}

solutions:

{{{x = (-4 +4.47 )/(-2) }}}

{{{x = 0.47 /-2 }}}

{{{x = -0.235 }}}..........first {{{x-intercept}}} is at point ({{{-0.235 }}},{{{0}}})

or

{{{x = (-4 -4.47 )/(-2) }}}

{{{x = -8.47 /-2 }}}

{{{x =4.235 }}}..............second {{{x-intercept}}} is at point ({{{4.235 }}},{{{0}}})




{{{ graph( 600, 600, -6, 6, -6, 6, -x^2+4x+1) }}}