document.write( "Question 887420: Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5 \n" ); document.write( "
Algebra.Com's Answer #536606 by nerdybill(7384)\"\" \"About 
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Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5
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\n" ); document.write( "y-value when x=1.5
\n" ); document.write( "y=3x^2+0.6x-3
\n" ); document.write( "y=3(1.5)^2+0.6(1.5)-3
\n" ); document.write( "y=6.75+0.9-3
\n" ); document.write( "y=4.65
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\n" ); document.write( "Now, you have a point at (1.5, 4.65)
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\n" ); document.write( "Next find the first derivative:
\n" ); document.write( "y' = 6x + 0.6
\n" ); document.write( "slope at x=1.5 is
\n" ); document.write( "6(1.5) + 0.6
\n" ); document.write( "9.6 (slope)
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\n" ); document.write( "Now, using the slope (9.6) and the single point (1.5, 4.65) plug into \"point-slope\" form:
\n" ); document.write( "y - y1 = m(x - x1)
\n" ); document.write( "y - 4.65 = 9.6(x - 1.5)
\n" ); document.write( "y - 4.65 = 9.6x - 14.4
\n" ); document.write( "y = 9.6x - 9.75 (equation of line in \"slope-intercept\" form)
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