Question 113934
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=2x^2+8x+9}}} we can see that a=2 and b=8


{{{x=(-8)/(2*2)}}} Plug in b=8 and a=2



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




{{{x=-2}}} Reduce



So the axis of symmetry is  {{{x=-2}}}



So the x-coordinate of the vertex is {{{x=-2}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(-2)}}}


{{{f(x)=2x^2+8x+9}}} Start with the given polynomial



{{{f(-2)=2(-2)^2+8(-2)+9}}} Plug in {{{x=-2}}}



{{{f(-2)=2(4)+8(-2)+9}}} Raise -2 to the second power to get 4



{{{f(-2)=8+8(-2)+9}}} Multiply 2 by 4 to get 8



{{{f(-2)=8+-16+9}}} Multiply 8 by -2 to get -16



{{{f(-2)=1}}} Now combine like terms



So the vertex is (-2,1)



Notice if you graph the equation {{{y=2x^2+8x+9}}} you can see that the vertex is (-2,1). So this visually verifies our answer.

{{{drawing(900,900,-12,8,-9,11,
grid( 1 ),
graph(900,900,-12,8,-9,11, 2x^2+8x+9),
circle(-2,1,0.05),
circle(-2,1,0.08)
)}}}