Point (p, 4p²) on Curve y = 4x² for All p

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The point (p, 4p²) lies on the curve defined by the equation y = 4x² for all real values of p. By substituting x = p into the equation, the corresponding y-coordinate is calculated as y = 4p², confirming that the point satisfies the equation of the curve. This demonstrates that the relationship holds true universally for any real number p, validating the assertion made in the discussion.

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Show that the point [tex](p, 4p^{2})[/tex] lies on the curve [tex]y = 4x^{2}[/tex] for all real values of p.

I'm not at all sure how to go about answering this. I know that substituting p into [tex]y = 4x^{2}[/tex] satisfies the equation. Is that enough?, or I'm not looking at this deep enough. Any hint(s) appreciated.

Thanks.
 
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Saying a point is on the graph given by some equation, you are saying the x- and y-coordinates satisfy that equation. So yes, since substituting x = p into your equation gives y = 4p^2, you know that [itex](p, 4p^2)[/itex] is on the graph.
 
Looks like the question wasn't as difficult as I thought it was going to be.

Okay statdad thanks. :)

P.S. Appologies to moderators. I'll post these sorts of questions in the correct section next time. ;)
 

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