SOLUTION: Where is the vertex of the graph of y=x^2-5. (resubmitted the first person that answered this didn't give the answer)

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Question 1709: Where is the vertex of the graph of y=x^2-5. (resubmitted the first person that answered this didn't give the answer)
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
quadratics are symmetric about the minimum or maximum point...their shape is either a U-shape (for a positive x^2 term) or is n-shaped (for a negative x^2 term).
The beauty and ease of finding a quadratic is "fun".
just pick some "obvious" points...like
1. when x=0, then y=-5
2. when y=0, then x%5E2=5, so x HAS 2 SOLUTIONS....+sqrt%285%29 and -sqrt%285%29.
So, we can instantly sketch the graph, and as the curve crosses the x-axis at the "same" point on both side of the y-axis, then it is a symmetric shape on the y-axis...so, the minimum point is at y=-5
graph%28300%2C200%2C-3%2C+3%2C+-6%2C+4%2C+%28x%5E2-5%29%29
cheers
Jon.