document.write( "Question 350591: Suppose that A, B, C, and D are constants and f is the cubic polynomial f(x)=Ax³+Bx²+Cx+D. Suppose also that the tangent line to y = f(x) at x = 0 is y = x and the tangent line at x = 2 is given by y = 2x – 3. Find the values of A, B, C, and D. Then sketch the graph of y = f(x) and the two tangent lines for -2 ≤ x ≤ 4. \n" ); document.write( "
Algebra.Com's Answer #250583 by Fombitz(32388)\"\" \"About 
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\"y=Ax%5E3%2BBx%5E2%2BCx%2BD\"
\n" ); document.write( "The slope of the tangent line at a given point is the value of the derivative at that point.
\n" ); document.write( "The derivative of the function is,
\n" ); document.write( "\"dy%2Fdx=3Ax%5E2%2B2Bx%2BC\"
\n" ); document.write( "At x=0, the slope of the tangent line is \"m=1\".
\n" ); document.write( "\"m=dy%2Fdx%280%29=3A%280%29%2B2B%280%29%2BC=1\"
\n" ); document.write( "\"highlight%28C=1%29\"
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\n" ); document.write( "At x=2, the slope of the tangent line is \"m=2\".
\n" ); document.write( "\"m=dy%2Fdx%282%29=3A%282%29%5E2%2B2B%282%29%2B1=2\"
\n" ); document.write( "\"12A%2B4B%2B1=2\"
\n" ); document.write( "1.\"12A%2B4B=1\"
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\n" ); document.write( "Using the tangent line at x=0, you also know the intersection point because,
\n" ); document.write( "\"y=x\"
\n" ); document.write( "\"y=0\"
\n" ); document.write( "(0,0) is also a point on the cubic.
\n" ); document.write( "\"y=Ax%5E3%2BBx%5E2%2BCx%2BD\"
\n" ); document.write( "\"y=A%280%29%2BB%280%29%2B%280%29%2BD=0\"
\n" ); document.write( "\"highlight%28D=0%29\"
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\n" ); document.write( "Similarly at x=2,
\n" ); document.write( "\"y=2x-3\"
\n" ); document.write( "\"y=2%282%29-3\"
\n" ); document.write( "\"y=1\"
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\n" ); document.write( "\"y=Ax%5E3%2BBx%5E2%2Bx\"
\n" ); document.write( "\"1=A%282%29%5E3%2BB%282%29%5E2%2B2\"
\n" ); document.write( "\"1=8A%2B4B%2B2\"
\n" ); document.write( "2.\"8A%2B4B=-1\"
\n" ); document.write( "Subtract eq. 2 from eq. 1,
\n" ); document.write( "\"12A%2B4B-8A-4B=1-%28-1%29\"
\n" ); document.write( "\"4A=2\"
\n" ); document.write( "\"highlight%28A=1%2F2%29\"
\n" ); document.write( "Then from eq. 2,
\n" ); document.write( "\"8%281%2F2%29%2B4B=-1\"
\n" ); document.write( "\"4%2B4B=-1\"
\n" ); document.write( "\"4B=-5\"
\n" ); document.write( "\"highlight%28B=-5%2F4%29\"
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