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conCentrate9
United States438 Posts
For the parabola y = x^2 + x find the equations of both tangent lines that pass through the point (2, -3). I started by finding the derivative of y as y' = 2x+1 and finding y'(2)=5, but using that slope and the point (2, -3) I get for my tangent line y = 5x - 13, which isn't even close. I just need direction I guess. Thanks.   | ||
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paper
13196 Posts
just make an equation from those two pieces of information >_> | ||
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conCentrate9
United States438 Posts
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paper
13196 Posts
lemme see | ||
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Dead9
United States4725 Posts
y + 3 = (2x+ 1)(x-2) y + 3 = 2x^2 - 3x - 2 y = 2x^2 - 3x - 5 2x^2 - 3x - 5 = x^2 + x x^2 - 4x - 5 = 0 (x - 5)(x + 1) I think you got it from here? | ||
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BlackStar
Netherlands3029 Posts
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paper
13196 Posts
basically, you want the slope of Y at the tangential point to equal the slope of the line from (2, -3) to that point. you know the eqn for the slope of Y [or 2x+1], and you find the slope from the point to the tangent point with (y+3)/(x-2) [which you get from the (2, -3)], so you set them equal. y+3 / x-2 = 2x + 1, and you know that y = x^2 + x, so you substitute it. solving for X gives you 5 and -1, which are the x-coordinates of where the tangent lines hit Y. | ||
micronesia
United States24747 Posts
+ Show Spoiler + This has been proposed and dismissed several times on the grounds that tl has nothing to do with hw and should not give that topic special treatment. However, nowadays pretty much all hw help is allowed via blogs, yet I think it should just be separate from blogs to relieve some of the burden on the blogs section. A small link somewhere to a hw section along with a rule that you can't request hw help in the blogs section would do two things. One is, it would give people a place to request hw help that wasn't cumbersome, and two it would discourage people a bit from asking for help since the hw help section would no doubt receive less attention. That or moderate hw help to some extent. | ||
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AngryLlama
United States1227 Posts
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conCentrate9
United States438 Posts
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