SOLUTION: Once again, I am in need of your help. I am stuck on a quadratic equation that I have to graph... and out of 16 problems on this worksheet, this is the only one I have not been abl

Question 136458: Once again, I am in need of your help. I am stuck on a quadratic equation that I have to graph... and out of 16 problems on this worksheet, this is the only one I have not been able to solve. This is not from my book, this is off of a worksheet.
This is what I have so far:
y= x^2 + 2
a=1 b=0 c=2
a=1 > 0 parabola opens upwards
x=0
y=0^2 + 2
y-intercept is (0,2)
y=0 0=x^2 +2
x= -0�√(0^2-4(1)(2)
-------------------------------
2(-1)

x= 0�√(0-8)
-----------------
2
X=0 and this is where I am stuck at. I am sorry your page is not allowing me to use the proper quadratic equation form. I did the best that I could.
Would you please help me work through this? Thanks!! ~Lynn

Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Lynn, your work so far is great. The thing you need to understand is that there may, in fact, be no real solution(s). If the determinant ends up being negative, then the solutions to the quadratic equation are complex. That means there will be no x value(s) that fall on the x axis.
So let's pick up where you left off.

These are complex numbers and won't plot on a 'standard' xy grid
So, how do you plot this parabola? You have one point (the vertex) and the basic shape. You know the graph will be symmetric with respect to a line through the vertex. In this case, given the shape, that line is the y axis.
So, you need a couple of points where x is not 0. Let's use x = 2. Then we can reflect that to x=-2. at x=2 becomes =
Plot points (2,6) and (-2,6), then draw in the curve as follows.

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