SOLUTION: how do you get the x intercepts by evaluating the function in order to get other points to graph the function>?????? this is what I have so far: f(x) =2x^2 +2 =2(x^2 +1)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: how do you get the x intercepts by evaluating the function in order to get other points to graph the function>?????? this is what I have so far: f(x) =2x^2 +2 =2(x^2 +1)       Log On


   

Question 313133: how do you get the x intercepts by evaluating the function in order to get other points to graph the function>??????
this is what I have so far:
f(x) =2x^2 +2
=2(x^2 +1) +2
f(x) = 2(x^2+1+.5-.5)+2
=2(x+2+1)+3(.5)+2
=2(x+3)^2 + 3.5

am i on the right track and if so what next in order to finish ??
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
No, not on the right track.
To get x-intercepts, set f(x)=y=0 and solve for x.
0=2x%5E2+%2B2
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%280%29%5E2%2B2
y=2
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You can find the axis of symmetry and vertex by completing the square.
y=2x%5E2%2B2
Actually, it's already in vertex form,
y=2%28x-0%29%5E2%2B2
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+%2B-+1, y=2%281%29%5E2%2B2=4
x=0+%2B-+2, y=2%282%29%5E2%2B2=10
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