Question 287314
The easiest way (if you don't have a graphing calculator or graphing software) is to turn it from general form (y=ax^2+bx+c) into standard parabola form (y=a(x-p)^2 + q)

You can do this by completing the square:

{{{y=2x^2+8x+4}}}

Factor out the 2 from the x terms:

{{{y=2(x^2+4x)+4}}}

Take half of the b=4 term, then square it {{{(4/2)^2 = 4}}}, and add it inside the brackets:

{{{y=2(x^2+4x+4)+4}}}

You added 8 (2 outside the brackets, times the 4 inside) to the equation, so you now have to subtract 8 from the outside:

{{{y=2(x^2+4x+4)+4-8}}}


Also note that the terms in the brackets now make a perfect square. {{{x^2+4x+4=(x+2)^2}}}


{{{y=2(x+2)^2-4}}}

This is a parabola with a=2, and center at (p,q)=(-2,-4)

If you want to see an image of this graph, or require more explanation, send me a message :)