Question 144744
Everything is correct except for the vertex



To find the vertex, we first need to find the axis of symmetry (ie the x-coordinate of the vertex)

To find the axis of symmetry, use this formula:


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


From the equation {{{y=x^2-2x-8}}} we can see that a=1 and b=-2


{{{x=(--2)/(2*1)}}} Plug in b=-2 and a=1



{{{x=2/(2*1)}}} Negate -2 to get 2



{{{x=(2)/2}}} Multiply 2 and 1 to get 2




{{{x=1}}} Reduce



So the axis of symmetry is  {{{x=1}}}



So the x-coordinate of the vertex is {{{x=1}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(1)}}}


{{{f(x)=x^2-2x-8}}} Start with the given polynomial



{{{f(1)=(1)^2-2(1)-8}}} Plug in {{{x=1}}}



{{{f(1)=(1)-2(1)-8}}} Raise 1 to the second power to get 1



{{{f(1)=-9}}} Now combine like terms



So the vertex is (1,-9)




So let's use the this  (along with the other pieces you solved for) to graph {{{y=x^2-2x-8}}}

{{{drawing(500,500,-9,11,-12,5,
grid( 1 ),
graph(500,500,-9,11,-12,5, x^2-2x-8)
)}}} Graph of {{{y=x^2-2x-8}}}