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 one year 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
 one year 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
 one year 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)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the derivative is \(f'(x)=26x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((1,23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1oops, i meant the derivative is \(f'(x)=2+6x\)

pottersheep
 one year 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?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1slope is \(\frac{y_2y_1}{x_2x_1}\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1here you have the point \((1,23)\) so the denominator is \(x+1\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1it says that \((1,23)\) is a point on the tangent line, so it is given to you

pottersheep
 one year 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
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1a generic point on the curve looks lik e \[(x,3+2x+3x^2)\] you also have the point \((1,23)\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1and 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much :)

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