document.write( "Question 1153266: Please help me\r
\n" ); document.write( "\n" ); document.write( "If the tangent line to y = f(x) at (4, 3) passes through the point (0, 2), find f(4) and f '(4).\r
\n" ); document.write( "\n" ); document.write( "f(4) =
\n" ); document.write( "f '(4) =
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Algebra.Com's Answer #775458 by MathLover1(20850) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "given:\r
\n" ); document.write( "\n" ); document.write( "the tangent line to $\"y+=+f%28x%29\"$ at ( $\"4\"$ , $\"3\"$ ) passes through the point ( $\"0\"$ , $\"2\"$ )\r
\n" ); document.write( "\n" ); document.write( "find: $\"f%284%29\"$ and $\"f\"$ ' $\"%284%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "The question gives you $\"f%284%29\"$ already, because the point ( $\"4\"$ , $\"3\"$ ) is given.
\n" ); document.write( "When $\"x+=4\"$ , $\"y=f%28x%29=f%284%29=+3\"$ .\r
\n" ); document.write( "\n" ); document.write( "We can find $\"f\"$ ' $\"%284%29\"$ by finding the $\"gradient\"$ at the point $\"f%284%29\"$ ,which we can do because we know the tangent touches both ( $\"4\"$ , $\"3\"$ ) and ( $\"0\"$ , $\"2\"$ ).\r
\n" ); document.write( "\n" ); document.write( "The gradient of a line is given by rise over run, or the change in $\"y\"$ divided by the change in $\"x\"$ , or\r
\n" ); document.write( "\n" ); document.write( " $\"m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29\"$ \r
\n" ); document.write( "\n" ); document.write( "We know two points on the graph in the question,so effectively we know the two values we need for $\"y\"$ and $\"x\"$ each. Say that\r
\n" ); document.write( "\n" ); document.write( "( $\"0\"$ , $\"2\"$ )→ $\"x%5B1%5D=0\"$ , $\"y%5B1%5D=2\"$ \r
\n" ); document.write( "\n" ); document.write( "( $\"4\"$ , $\"3\"$ )→ $\"x%5B2%5D=4\"$ , $\"y%5B2%5D=3\"$ \r
\n" ); document.write( "\n" ); document.write( "so\r
\n" ); document.write( "\n" ); document.write( " $\"m=%283-2%29%2F%284-0%29=1%2F4\"$ which is the gradient or $\"f\"$ ' $\"%284%29\"$ \r
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