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pottersheep
 3 years ago
Please help. Calculus!
Find the point of tangency and the equation of the tangent line passing through the point (1.23) to the graph of the function f(x) = 3 + 2x + 3x^2
pottersheep
 3 years ago
Please help. Calculus! Find the point of tangency and the equation of the tangent line passing through the point (1.23) to the graph of the function f(x) = 3 + 2x + 3x^2

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pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0The answer says:q (4,43) y = 22x  45 AND y = 14x  9 at q(2,19)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the derivative is \(f'(x)=26x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((1,23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops, i meant the derivative is \(f'(x)=2+6x\)

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for your time. Sorry, why is it (2+2x+2x2+23)/x+1? Where did the x+1 come from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0slope is \(\frac{y_2y_1}{x_2x_1}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here you have the point \((1,23)\) so the denominator is \(x+1\)

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0What exactly does the point P(1,23) mean?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it says that \((1,23)\) is a point on the tangent line, so it is given to you

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0What does it give me? Is it because a tangent is a straight line so I can use that point and the slope formula?

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, I thought it was where the tangent met f(x) = 3 + 2x + 3x^2....but that didnt make sense to me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a generic point on the curve looks lik e \[(x,3+2x+3x^2)\] you also have the point \((1,23)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and you know that the line with this slope is tangent to the curve at some point the slope my therefore be equal to the derivative at that point, which is \(2+6x^2\)

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you so much :)

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0You are such a good teacher =D
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