Question 694630
{{{f(x)=-(1/3)x^2-2x-9}}} in the form {{{f(x)=a(x-h)^2+k}}}

*[invoke completing_the_square -1/3, -2, -9]

{{{ x-intercept}}} ......set {{{f(x)=0}}}

{{{0=-(1/3)x^2-2x-9}}}...use quadratic formula to solve for {{{x}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-(-2) +- sqrt((-2)^2-4*(-1/3)*(-9) ))/(2*(-1/3)) }}}


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


{{{x = (2 +- sqrt(-8 ))/(-2/3)) }}}


{{{x = (2 +- 2.83*i)/(-2/3)) }}}


{{{x = 3(2 +- 2.83*i)/-2 }}}


{{{x = (6 +- 8.49*i)/-2 }}}

{{{x = (6/-2 +- 8.49*i/-2) }}}


{{{x = -3 +- 4.245*i }}}....so, we have two complex solutions, it means there is no real solutions and there is no {{{ x-intercepts}}}


and {{{y-intercept}}}......set {{{x=0}}}

{{{f(x)=-(1/3)0^2-2*0-9}}}

{{{f(x)=0-0-9}}}

{{{f(x)=-9}}}.......{{{y-intercept}}} is at ({{{0}}},{{{-9}}})


see it on a graph:

{{{ graph( 600, 600, -10,10, -15, 10, -(1/3)x^2-2x-9) }}}


 domain:


*[tex \LARGE \{\forall\,x\,|\,x\,\in\,\mathbb{R}\}]


 range:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{y\,|\,y\,\in\,\mathbb{R},\,y\ \leq\ -6\}]