document.write( "Question 1165141: https://answers.yahoo.com/question/index?qid=20200917201613AAEazzn\r
\n" ); document.write( "\n" ); document.write( "Help with this problem
\n" ); document.write( "

Algebra.Com's Answer #789599 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "There are two transformations to get from \"y=x%5E2\" to \"y=%28x%2B2%29%5E2-3\".

\n" ); document.write( "To determine the order of the transformations, consider how you would evaluate the new function for a given x value: add 2, square it, then subtract 3.

\n" ); document.write( "So the adding 2 is the first transformation, and the subtracting 3 is the second.

\n" ); document.write( "The graph of \"y=x%5E2\" has it minimum value when x=0; that value is 0. So the vertex of the graph is at (0,0).

\n" ); document.write( "\"graph%28400%2C400%2C-4%2C4%2C-4%2C20%2Cx%5E2%29\"

\n" ); document.write( "The first transformation is from \"y=x%5E2\" to \"y=%28x%2B2%29%5E2\". That transformation moves the whole graph 2 units to the LEFT.

\n" ); document.write( "Moving the graph LEFT when the transformed equation is x PLUS 2 squared is confusing to many beginning students. But it makes perfect sense if you look at it this way:

\n" ); document.write( "The graph of \"y+=+%28x%2B2%29%5E2\" has its minimum value when x+2=0 -- but that is when x = -2.

\n" ); document.write( "So now the vertex is at (-2,0).

\n" ); document.write( "So the first transformation moves the whole graph 2 units to the left:

\n" ); document.write( "\"graph%28400%2C400%2C-4%2C4%2C-4%2C20%2Cx%5E2%2C%28x%2B2%29%5E2%29\"

\n" ); document.write( "The second transformation is from \"y=%28x%2B2%29%5E2\" to \"%28x%2B2%29%5E2-3\". It should be easy to understand that this transformation simply moves the whole graph down 3 units; so now the vertex is at (-2,-3):

\n" ); document.write( "\"graph%28400%2C400%2C-4%2C4%2C-4%2C20%2Cx%5E2%2C%28x%2B2%29%5E2%2C%28x%2B2%29%5E2-3%29\"

\n" ); document.write( "
\n" ); document.write( "
\n" );