What is the equation of the normal to y= (x^2 + 3) / (x+1) at (1,2)? the answer is x =1 but I don't get why. I know that the slope is 0 (after substituting 1 into the derivative equation), so to find the normal of that slope, isn't it 0 also? I ended up with y=2, which is wrong..

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What is the equation of the normal to y= (x^2 + 3) / (x+1) at (1,2)? the answer is x =1 but I don't get why. I know that the slope is 0 (after substituting 1 into the derivative equation), so to find the normal of that slope, isn't it 0 also? I ended up with y=2, which is wrong..

Mathematics
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Normal meas perpendicular. If the slope of one line is 0, then the slope of the normal must be undefined (vertical, in other words).
(The use of the word goes back to lenses. A "normal" ray of light hits the glass of the lens at right angles to the surface. A normal ray is not bent by the lens, at least on entry.)
I know it's vertical but how do you find the vertical slope? :S

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Okay, you found the derivative and plugged in x=1 to get the slope, 0/2. Since "normal" means perpendicular, you want a slope that is perpendicular to 0/2. Two slopes are perpendicular if they are negative reciprocals of each other, like 3/5 and -5/3. In this case, the neg. recip. is -2/0. That's undefined, and vertical lines have an undefined slope. Does that help?
Finally, since the vertical line passes through (1, 2), its equation must be x = 1.
Oh I get it. Thanks so much Bickford :)

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