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At what point does the normal line to $y=-5+4x+3x^2$ at $(1,2)$ intersect the parabola a second time?

$y'=6x+4$
$m_{tangent}=6(1)+4=10$
$m_{normal}=-\dfrac{1}{10}$

$y=f'(1)(x-1)+f(1)$
$=(-1/10)(x-1)+2$
$y=-1/10+21/10 \implies$ equation of normal line

Then, set normal line = curve to find point of intersection: $(-1/10x+21/10=-5+4x+3x^2)(-10) \rightarrow$ I multiplied by $-10$ to get rid of the $-1/10x$ fraction.

$x=60-4x-30x^2$
$-5(6x^2+x-12)=0$

Then I used the quadratic formula to find the $x$ values. I got $x= \dfrac{4}{3},-\dfrac{3}{2}$ and this is wrong.

How do I find the $x$ values and $y$ values and what am I doing wrong? Thanks.

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    $\begingroup$ Well, one of the points of intersection you got should have been (1,2), so that's your first sign something's gone wrong. $\endgroup$ Sep 23 '14 at 5:16
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Correcting some algebra, will look into the analytic geometry when I have time.

$$C_{norm}: y= \frac{21}{10}-\frac{x}{10}$$

Then you have $$\frac{21}{10}-\frac{x}{10}=-5+4x+3x^2 \iff -3x^2-\frac{41x}{10}+\frac{71}{10}=0 \iff -\frac{1}{10} ((x-1)(71+30x))=0 \iff (x-1)(71+30x)=0 \iff x=1 \; \mathrm{or} \; x=-\frac{71}{30}$$

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  • $\begingroup$ +1 great, thanks so much. You forgot a negative sign in front of the $5$ though. $\endgroup$
    – Emi Matro
    Sep 23 '14 at 5:59
  • $\begingroup$ Don't worry, it was a typo, I considered it $-5$ in the rest of the solution. Thanks. $\endgroup$
    – UserX
    Sep 23 '14 at 6:01

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