Is either x=9 or x=15 a solution to x-3=12?
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Okay so I attatched pictures of my work. I would just graph it by plugging in x-values on either side of the vertex into the equation. For example, the vertex is at (-2,3). So try plugging x=-5 and x=1 into the equation (y=a(x-h)^2 +k).
y=(x+2)^2 +3
y=((1)+2)^2 +3=(3)^2+3= 9+3= 12
Now you can plot the vertex (-2,3), the point (1,12), and (-5,12). From there just connect the dots and you should get a graph that looks similar to the last picture I attached. Hope this all made sense sorry it was so wordy I just wasn't sure how much you knew or didn't know. So, I thought it would be better to explain it all rather than not enough.
y=(x+2)^2 +3
y=((1)+2)^2 +3=(3)^2+3= 9+3= 12
Now you can plot the vertex (-2,3), the point (1,12), and (-5,12). From there just connect the dots and you should get a graph that looks similar to the last picture I attached. Hope this all made sense sorry it was so wordy I just wasn't sure how much you knew or didn't know. So, I thought it would be better to explain it all rather than not enough.