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The answer says:q (-4,43) y = -22x - 45 AND y = 14x - 9 at q(2,19)

set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

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

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 \[\frac{3+2x+2x^2+23}{x+1}\]

I get it now!!!

Thank you so much :)

ok good
yw

You are such a good teacher =D