Question 103996
To find the axis of symmetry, use this formula:


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


From the equation {{{y=2x^2+4x-5}}} we can see that a=2 and b=4


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



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




{{{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)=2x^2+4x-5}}} Start with the given polynomial



{{{f(-1)=2(-1)^2+4(-1)-5}}} Plug in {{{x=-1}}}



{{{f(-1)=2(1)+4(-1)-5}}} Raise -1 to the second power to get 1



{{{f(-1)=2+4(-1)-5}}} Multiply 2 by 1 to get 2



{{{f(-1)=2+-4-5}}} Multiply 4 by -1 to get -4



{{{f(-1)=-7}}} Now combine like terms



So the vertex is (-1,-7)




Notice if you graph the equation {{{y=2x^2+4x-5}}} you get

{{{drawing(900,900,-11,9,-17,3,
grid( 1 ),
graph(900,900,-11,9,-17,3, 2x^2+4x-5),
circle(-3,1,0.05),
circle(-3,1,0.08),
circle(-2,-5,0.05),
circle(-2,-5,0.08),
circle(-1,-7,0.05),
circle(-1,-7,0.08),
circle(0,-5,0.05),
circle(0,-5,0.08),
circle(1,1,0.05),
circle(1,1,0.08)
)}}}



and you can see that the vertex is (-1,-7)