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he66666
 5 years ago
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..
he66666
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Normal meas perpendicular. If the slope of one line is 0, then the slope of the normal must be undefined (vertical, in other words).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(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.)

he66666
 5 years ago
Best ResponseYou've already chosen the best response.0I know it's vertical but how do you find the vertical slope? :S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Finally, since the vertical line passes through (1, 2), its equation must be x = 1.

he66666
 5 years ago
Best ResponseYou've already chosen the best response.0Oh I get it. Thanks so much Bickford :)
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