Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (moongazer):

For what values of c is the curve y = c/(x+1) tangent to the line through the points (0,3) and (5, -2)?

OpenStudy (moongazer):

help me please :)

OpenStudy (unklerhaukus):

have you found the equation of the line through those two points?

OpenStudy (moongazer):

is it y = -x + 3? this is for the equation for the tangent line. am I correct?

OpenStudy (moongazer):

@UnkleRhaukus

OpenStudy (unklerhaukus):

yes that is the line through the two points

OpenStudy (moongazer):

what should I do next?

OpenStudy (moongazer):

@UnkleRhaukus

OpenStudy (unklerhaukus):

when do the equations y = c/(x+1) y = -x + 3 meet?

OpenStudy (unklerhaukus):

i think you have to find the value for c which makes one solution to those equations

OpenStudy (moongazer):

ohh, so I just need to equate those two and solve for c? am I correct?

OpenStudy (unklerhaukus):

well equate them yes, c/(x+1) = -x + 3 what so can simply it to

OpenStudy (moongazer):

c = (-x + 3)(x+1) is that it?

OpenStudy (moongazer):

Why did my teacher got the derivative of y = c/(x+1)? He didn't finish the solution, so I don't know if i'm right.

OpenStudy (unklerhaukus):

you want there to be exactly one solution, you'll have to expand the brackets and complete the square,

OpenStudy (unklerhaukus):

oh, we can use derivatives?

OpenStudy (moongazer):

Our topic is differential calculus , so I think I need to use derivatives here. yes

OpenStudy (unklerhaukus):

oh right sorry, i was thinking of a different approach.

OpenStudy (unklerhaukus):

well can you take the derivative of y = c/(x+1)

OpenStudy (unklerhaukus):

\[\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]

OpenStudy (moongazer):

I got y' = -c/(x+1)^2

OpenStudy (unklerhaukus):

thats right

OpenStudy (moongazer):

what should I do next?

OpenStudy (moongazer):

This is were I'm stuck :)

OpenStudy (unklerhaukus):

i dont really understand this question

OpenStudy (moongazer):

ohhh

OpenStudy (moongazer):

I think the next step is -c/(x+1)^2 = -1 because the derivative of y = c/(x+1) is it's tangent line

OpenStudy (unklerhaukus):

yeah, .,

OpenStudy (anonymous):

How about this, using discriminants? y = c / (x + 1) y = -x + 3 -x + 3 = c / (x + 1) (-x + 3)(x + 1) = c -x^2 + 2x + 3 = c x^2 - 2x - 3 = -c x^2 - 2x + c - 3 = 0 b^2 - 4ac = 0 (-2)^2 - 4(1)(c - 3) = 0 4 - 4c + 12 = 0 4c = 16 c = 4

OpenStudy (moongazer):

@TURITW Why is it that b^2 - 4ac = 0 ?

OpenStudy (anonymous):

b^2 - 4ac > 0 means 2 real and distinct roots. b^2 - 4ac = 0 means 2 real and equal roots. b^2 - 4ac < 0 means 2 complex roots.

OpenStudy (moongazer):

Why do you need to have equal roots?

OpenStudy (anonymous):

Equal roots means tangent, because the curve needs to touch the line once

OpenStudy (moongazer):

Why is it that equal roots means tangent? I'm sorry if it's a dumb question. :)

OpenStudy (moongazer):

@TURITW

OpenStudy (anonymous):

Equal roots means same coordinates, so it means it only cuts the line once, example, It cuts the line at (1 , 1) (just a random coordinates) only.

OpenStudy (moongazer):

Thanks, I'll just show my teacher two different solutions. I'll just ask for the correct answer. :)

OpenStudy (moongazer):

Thanks to both of you. :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!