Question

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

Answer 1

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

Answer 2

the derivative is \(f'(x)=2-6x\)
the slope of the tangent line through \((x,2+2x+3x^2)\) and \((-1,-23)\) is
\[\frac{2+2x+2x^2+23}{x+1}\]

Answer 3

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

Answer 4

oops, i meant the derivative is \(f'(x)=2+6x\)

Answer 5

Thank you for your time. Sorry, why is it (2+2x+2x2+23)/x+1? Where did the x+1 come from?

Answer 6

slope is \(\frac{y_2-y_1}{x_2-x_1}\)

Answer 7

here you have the point \((-1,-23)\) so the denominator is \(x+1\)

Answer 8

What exactly does the point P(-1,-23) mean?

Answer 9

it says that \((-1,-23)\) is a point on the tangent line, so it is given to you

Answer 10

What does it give me? Is it because a tangent is a straight line so I can use that point and the slope formula?

Answer 11

Ah, I thought it was where the tangent met f(x) = 3 + 2x + 3x^2....but that didnt make sense to me

Answer 12

a generic point on the curve looks lik e
\[(x,3+2x+3x^2)\] you also have the point \((-1,-23)\)

Answer 13

the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]

Answer 14

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 \(2+6x^2\)

Answer 15

I get it now!!!

Answer 16

Thank you so much :)

Answer 17

ok good
yw

Answer 18

You are such a good teacher =D