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January 31, 2015

January 31, 2015

Posted by **Shelly** on Monday, October 3, 2011 at 8:18pm.

Set up an equation, which when solved will give the points where the tangent line has y-intercept 5.

- calculus -
**Damon**, Monday, October 3, 2011 at 8:40pmf' = a = -8(2x+1)/(x^2+x+2)^2

y = a x + 5

8/(x^2+x+2) = -8(2x^2+x)/ (x^2+x+2)^2 + 5

etc

- calculus -
**Shelly**, Monday, October 3, 2011 at 8:46pmDamon, so what you did was apply the quotient rule to the ax part? Wouldn't the -8 be -8x then? And if I'm doing it right, would my final equation be this?

0=(5x^4+10x^3+x^2+4x+4)/(x^2+x+2)^2

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