Question 341980
You are correct. The value of b is {{{b=0}}} since {{{ y= -6x^2-1 }}} is the same as {{{ y= -6x^2+0x-1 }}}



So the x coordinate of the vertex is {{{x=-b/2a=-0/(2(-6))=0}}}



Plug this back in to get {{{y=-6(0)^2-1=-1}}}. So the vertex is (0, -1)



So you are correct. The max value is -1.



To solve {{{ y= -6x^2-1 }}}, plug in y=0 to get {{{ 0= -6x^2-1 }}}




Notice that the quadratic {{{-6x^2-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=-6}}}, {{{B=0}}}, and {{{C=-1}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = ((0) +- sqrt( (0)^2-4(-6)(-1) ))/(2(-6))}}} Plug in  {{{A=-6}}}, {{{B=0}}}, and {{{C=-1}}}



{{{x = (0 +- sqrt( 0-4(-6)(-1) ))/(2(-6))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0-24 ))/(2(-6))}}} Multiply {{{4(-6)(-1)}}} to get {{{24}}}



{{{x = (0 +- sqrt( -24 ))/(2(-6))}}} Subtract {{{24}}} from {{{0}}} to get {{{-24}}}



{{{x = (0 +- sqrt( -24 ))/(-12)}}} Multiply {{{2}}} and {{{-6}}} to get {{{-12}}}. 



{{{x = (0 +- 2i*sqrt(6))/(-12)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (0+2i*sqrt(6))/(-12)}}} or {{{x = (-0-2i*sqrt(6))/(-12)}}} Break up the expression.  



{{{x = -(i*sqrt(6))/(6)}}} or {{{x = (i*sqrt(6))/(6)}}} Reduce and simplify



So the solutions are {{{x = -(i*sqrt(6))/(6)}}} or {{{x = (i*sqrt(6))/(6)}}}



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Jim