Question 109691
To find the vertex, we need to know the axis of symmetry


To find the axis of symmetry, use this formula:


{{{x=-b/(2a)}}}


From the equation {{{y=-4x^2+40x-93}}} we can see that a=-4 and b=40


{{{x=(-40)/(2*-4)}}} Plug in b=40 and a=-4



{{{x=(-40)/-8}}} Multiply 2 and -4 to get -8




{{{x=5}}} Reduce



So the axis of symmetry is  {{{x=5}}}



So the x-coordinate of the vertex is {{{x=5}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(5)}}}


{{{f(x)=-4x^2+40x-93}}} Start with the given polynomial



{{{f(5)=-4(5)^2+40(5)-93}}} Plug in {{{x=5}}}



{{{f(5)=-4(25)+40(5)-93}}} Raise 5 to the second power to get 25



{{{f(5)=-100+40(5)-93}}} Multiply 4 by 25 to get 100



{{{f(5)=-100+200-93}}} Multiply 40 by 5 to get 200



{{{f(5)=7}}} Now combine like terms



So the vertex is (5,7)




Notice if we graph the equation {{{y=-4x^2+40x-93}}} we get

{{{drawing(900,900,-5,15,-3,17,
grid( 1 ),
graph(900,900,-5,15,-3,17, -4x^2+40x-93),
circle(3,-9,0.05),
circle(3,-9,0.08),
circle(4,3,0.05),
circle(4,3,0.08),
circle(5,7,0.05),
circle(5,7,0.08),
circle(6,3,0.05),
circle(6,3,0.08),
circle(7,-9,0.05),
circle(7,-9,0.08)
)}}}

and we can see that the vertex is (5,7)