SOLUTION: (x-4)^2+16(y-5)=0 Does this line go through point (0,7)? And does this parabola open upwards?

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Question 546005: (x-4)^2+16(y-5)=0 Does this line go through point (0,7)? And does this parabola open upwards?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
(x-4)^2+16(y-5)=0 Does this line go through point (0,7)? And does this parabola open upwards?
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Standard form of an equation for a parabola that opens downwards: (x-h)^2=-4p(y-k), with (h,k) being the (x,y) coordinates of the vertex.
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Rewrite given equation:
(x-4)^2=-16(y-5)
This is the standard form of an equation for a parabola that opens downwards with vertex at (4,5)
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plug (0,7) into equation:
(0-4)^2=-16(7-5)
16≠-32
Therefore, (0,7) is not a point on the parabola