document.write( "Question 41253: Can you please explain to me how you create an equation (function) from looking at a parabola? I can make a parabola from an equation. But not the other way around. I need very easy and explained steps!
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Algebra.Com's Answer #26656 by psbhowmick(878)\"\" \"About 
You can put this solution on YOUR website!
Look at the parabola below.
\n" ); document.write( "\"graph%28+200%2C300%2C-1%2C7%2C-1%2C19%2C2%2B%28x-3%29%5E2%2C2+%29\"\r
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\n" ); document.write( "\n" ); document.write( "At first find the vertex of the parabola from the figure.
\n" ); document.write( "You can easily do this.
\n" ); document.write( "Here the coordinates of the vertex are (3,2).
\n" ); document.write( "Also you can clearly see that the parabola is having an axis parallel to the x-axis.\r
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\n" ); document.write( "\n" ); document.write( "If the coordinates of the vertex of a parabola with axis parallel to x-axis be (h,k) then its equation is of the form
\n" ); document.write( "\"y-k+=+4%2Aa%2A%28x-h%29%5E2\"__________________(1)\r
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\n" ); document.write( "\n" ); document.write( "In this case h = 3, k = 2
\n" ); document.write( "Put this values in equation (1) then you get
\n" ); document.write( "\"y-2+=+4%2Aa%2A%28x-3%29%5E2\"__________________(2)\r
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\n" ); document.write( "\n" ); document.write( "Now, note the coordinates of the point where the parabola cuts one of the coordinate axes.
\n" ); document.write( "Here the parabola cuts the y-axis at (0,11).
\n" ); document.write( "So (0,11) must be a point on the parabola and hence must satisfy its equation.
\n" ); document.write( "Hence x=0 and y=11 satisfies equation (2).
\n" ); document.write( "Therefore, \"11-2+=+4%2Aa%2A%280-3%29%5E2\"
\n" ); document.write( "or \"9+=+4%2Aa%2A3%5E2+=+36a\"
\n" ); document.write( "or \"a+=+1%2F4\"
\n" ); document.write( "Put \"a+=+1%2F4\" in equation and get the required equation of the given parabola.
\n" ); document.write( "Here, the equation of the given parabola is \"y-2+=+%28x-3%29%5E2\"
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