pottersheep
  • pottersheep
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
pottersheep
  • pottersheep
The answer says:q (-4,43) y = -22x - 45 AND y = 14x - 9 at q(2,19)
anonymous
  • anonymous
the derivative is \(f'(x)=2-6x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((-1,-23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]
anonymous
  • anonymous
set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.