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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

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The answer says:q (-4,43) y = -22x - 45 AND y = 14x - 9 at q(2,19)
the derivative is \(f'(x)=2-6x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((-1,-23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]
set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

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Other answers:

oops, i meant the derivative is \(f'(x)=2+6x\)
Thank you for your time. Sorry, why is it (2+2x+2x2+23)/x+1? Where did the x+1 come from?
slope is \(\frac{y_2-y_1}{x_2-x_1}\)
here you have the point \((-1,-23)\) so the denominator is \(x+1\)
What exactly does the point P(-1,-23) mean?
it says that \((-1,-23)\) is a point on the tangent line, so it is given to you
What does it give me? Is it because a tangent is a straight line so I can use that point and the slope formula?
Ah, I thought it was where the tangent met f(x) = 3 + 2x + 3x^2....but that didnt make sense to me
a generic point on the curve looks lik e \[(x,3+2x+3x^2)\] you also have the point \((-1,-23)\)
the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]
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 \(2+6x^2\)
I get it now!!!
Thank you so much :)
ok good yw
You are such a good teacher =D

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