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anonymous
 one year ago
Find the equation of the tangent line to f(x) = 9 − x^2 through the point (5, 0) that touches f (x) at a point below the xaxis.
anonymous
 one year ago
Find the equation of the tangent line to f(x) = 9 − x^2 through the point (5, 0) that touches f (x) at a point below the xaxis.

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dan815
 one year ago
Best ResponseYou've already chosen the best response.2first you need to find the slope at this point

dan815
 one year ago
Best ResponseYou've already chosen the best response.2that is the function of all the slopes wrt to x

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(x) is telling you all the heights at every x f'(x) is teelling you all the slopes at every x now at what x do we need to know our f'(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the point where it touches f(x) which is below the x axis?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2take a look at your point

dan815
 one year ago
Best ResponseYou've already chosen the best response.2whats the x value of the point

dan815
 one year ago
Best ResponseYou've already chosen the best response.2right so we need to know f'(5) = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But that point doesn't even lie on the curve...

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we just want the slope there at x=5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that would be 2(5) = 10, right?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh wait 5,0 isnt on this curve really?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah cause try plugging it in the function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we should first find a point that apparently lies somewhere under the xaxis

dan815
 one year ago
Best ResponseYou've already chosen the best response.2k lets graph and see what exactly is going on here

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay so eventually this has to happen, they want us to find this x,y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So are we solving for the secant line?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0between (5,0) and (a, 9a^2)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2you already got that the slope was F'(x)=2x

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now write the slope with those 2 points

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Any line through \((5,0)\) is of form \[y=m(x5)\tag{1}\] we want that line to "just touch" below parabola \[y=9x^2\tag{2}\] simply set both equations equal to each other and use the discriminant to find the slope, \(m\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2and equate it to the F'(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah lol i was typing the same xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We get a quadratic equation when we solve for a... that means we'll get 2 values for a...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, two tangent lines can be drawn to a parabola form an external point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol xD ok so a=1 and x=9 ? Is that right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435218241639:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2a=9 is right, but how did u figure ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but neither of the 2 points i found for 'a' lie under the xaxis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I solved the quadratic a^2 10a +9

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2okay i see you simply discarded a=1 without any reason

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nope i wrote that as well lol ^^ check again :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Okay just making sure :) plugin x=a=9 into the equation, you do get a point under the x axis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do? Ohh thats right! xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that means I get 2 tangent lines?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2plugin x=a=1, you get y = 9x^2 = 91 = 8 that means the tangent line touches the parabola at (1, 8) which is above x axis so discard this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I love discarding stuff in math xD haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lol we're very much alike :P

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i am not following exactly what u are doing here ganeshie y=m(x5) y=9x^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2are u then setting m=2x to solve it

dan815
 one year ago
Best ResponseYou've already chosen the best response.2arent there multiple solutions to inf solutions to m that intersect with this poiint on the parabola, how are you limiting to the case where its just touching

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2nope set both equations equal to each other and use discriminant : m(x5) = 9x^2 x^2 + mx  (5m+9) = 0 For this quadratic to have a single solution, the discriminant must be 0 : m^2 + 4(5m+9) = 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2solving that gives the slope of two tangent lines passing thru (5, 0)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh i see! what you mean i didnt understand what discriminant meant lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0neither did I o.o XD lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Lol that had me confused to but I was too scared to ask, nice one.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats okay, stick to dan's method as it is more calculussy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I feel like an idiot now xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No I totally understood everything we did!! I just didn't get what you were explaining to dan at the end xP

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics pshhhh in that case i'm more of an idiot that everyone! XD lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2basically what he did was, well normally to a line passing through that point (5,0) to your parabola there will be 2 intersections everywhere except for 1 place

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2reminds me of 3 idiots movie

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lmao XD truuueee :D so much humour the night before my exam :D lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2just to conclude : x=a = 9 gives slope=2x=2*9 = 18 so the equation of desired tangent line is y = 18(x5)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2right, and u know how from the quadratic equation the 2 differnet solutions come from that +/ part in squareroot

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much!!! :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Thank you teachers!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Are all these questions part of a practice exam?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, they're some sample/practice questions... or whatever they call it XD lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics :D sorry about the late response, wasn't paying attention lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Lol no worries was helping someone anyways, haha and good luck, calc exams have always been long >.<

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you! :) Yeah it's gonna be 2.5 hours 8:30 in the morning >.<

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Eek, haha get some sleep as well! Later :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'd better xD lol otherwise its a big fat 0 xD haha
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