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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
 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
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.1
the derivative is \(f'(x)=26x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((1,23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oops, i meant the derivative is \(f'(x)=2+6x\)
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.1
slope is \(\frac{y_2y_1}{x_2x_1}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
here you have the point \((1,23)\) so the denominator is \(x+1\)
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.1
it says that \((1,23)\) is a point on the tangent line, so it is given to you
 one year ago

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

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

satellite73Best ResponseYou've already chosen the best response.1
a generic point on the curve looks lik e \[(x,3+2x+3x^2)\] you also have the point \((1,23)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and 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 \(2+6x^2\)
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
Thank you so much :)
 one year ago

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