Question 313133
No, not on the right track.
To get x-intercepts, set f(x)=y=0 and solve for x.
{{{0=2x^2 +2}}}
Well there are no x such that when you square them and add 2 will give you zero. There are no x-intercepts for this parabola.
It never crosses the x-axis.
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To get the y-intercept, set x=0 and solve for y.
{{{y=2(0)^2+2}}}
{{{y=2}}}
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You can find the axis of symmetry and vertex by completing the square.
{{{y=2x^2+2}}}
Actually, it's already in vertex form,
{{{y=2(x-0)^2+2}}}
So the vertex is at (0,2) and it opens upwards. 
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To get other points, just pick an x and calculate y.
{{{x=0 +- 1}}}, {{{y=2(1)^2+2=4}}}
{{{x=0 +- 2}}}, {{{y=2(2)^2+2=10}}}
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{{{ drawing( 300, 300, -5, 5, -2, 12, grid(1),
circle(-1,4,.2),circle(0,2,.2),
circle(1,4,.2),
circle(-2,10,.2),
circle(2,10,.2),
graph( 300, 300, -5, 5, -2, 12,  2x^2+2)) }}}