document.write( "Question 1091721: Find the value of a and b so that the graph of y=sqrt(ax + b) passes through points (-4,6) and (1,4). \n" ); document.write( "
Algebra.Com's Answer #706170 by MathLover1(20849)\"\" \"About 
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\n" ); document.write( " In triangle ABC, median AM is such that angle BAC is divided in ratio 1:2. AM is extended through M to D so that angle DBA is a right angle, then the ratio AC : AD is equal to \r
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\n" ); document.write( "\n" ); document.write( "a median is a line joining a vertex with the mid-point of the opposite side\r
\n" ); document.write( "\n" ); document.write( "to find the value of \"a+\" and \"b\" so that the graph of \"y=sqrt%28ax+%2B+b%29\" use given points (\"-4\",\"6\") and (\"1\",\"4\")\r
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\n" ); document.write( "\n" ); document.write( "\"y=sqrt%28ax+%2B+b%29\" ...for (\"-4\",\"6\")\r
\n" ); document.write( "\n" ); document.write( "\"6=sqrt%28a%28-4%29+%2B+b%29\"
\n" ); document.write( "\"6=sqrt%28-4a+%2B+b%29\" ....square both sides\r
\n" ); document.write( "\n" ); document.write( "\"36=-4a+%2B+b\"...solve for \"b\"\r
\n" ); document.write( "\n" ); document.write( "\"36%2B4a+=+b\"....eq.1\r
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\n" ); document.write( "\n" ); document.write( "\"y=sqrt%28ax+%2B+b%29\" ...for (\"1\",\"4\")\r
\n" ); document.write( "\n" ); document.write( "\"4=sqrt%28a%281%29+%2B+b%29\"
\n" ); document.write( "\"4=sqrt%28a+%2B+b%29\" ....square both sides\r
\n" ); document.write( "\n" ); document.write( "\"16=a+%2B+b\"...solve for \"b\"\r
\n" ); document.write( "\n" ); document.write( "\"16-a+=+b\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "from eq.1 and eq.2 we have\r
\n" ); document.write( "\n" ); document.write( "\"36%2B4a+=+16-a+\"..solve for \"a\"\r
\n" ); document.write( "\n" ); document.write( "\"a%2B4a+=+16-36+\"\r
\n" ); document.write( "\n" ); document.write( "\"5a+=+-20+\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28a+=+-4+%29\"\r
\n" ); document.write( "\n" ); document.write( "now find \"b\"\r
\n" ); document.write( "\n" ); document.write( "\"16-%28-4%29+=+b\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "\"16%2B4+=+b\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28b=20%29\"\r
\n" ); document.write( "\n" ); document.write( "so, your equation is: \"y=sqrt%28-4x+%2B20%29\"\r
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