anonymous
  • anonymous
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 x-axis.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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dan815
  • dan815
first you need to find the slope at this point
anonymous
  • anonymous
-2x ?
dan815
  • dan815
that is the function of all the slopes wrt to x

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More answers

dan815
  • dan815
f(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
  • anonymous
the point where it touches f(x) which is below the x axis?
dan815
  • dan815
take a look at your point
dan815
  • dan815
(5, 0)
dan815
  • dan815
whats the x value of the point
anonymous
  • anonymous
5
dan815
  • dan815
right so we need to know f'(5) = ?
anonymous
  • anonymous
But that point doesn't even lie on the curve...
dan815
  • dan815
we just want the slope there at x=5
anonymous
  • anonymous
so that would be -2(5) = -10, right?
dan815
  • dan815
oh wait 5,0 isnt on this curve really?
dan815
  • dan815
u are right lol xD
anonymous
  • anonymous
Yeah cause try plugging it in the function
anonymous
  • anonymous
XD
anonymous
  • anonymous
I think we should first find a point that apparently lies somewhere under the x-axis
dan815
  • dan815
k lets graph and see what exactly is going on here
dan815
  • dan815
|dw:1435217659239:dw|
dan815
  • dan815
|dw:1435217751709:dw|
dan815
  • dan815
okay so eventually this has to happen, they want us to find this x,y
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
So are we solving for the secant line?
anonymous
  • anonymous
between (5,0) and (a, 9-a^2)
dan815
  • dan815
you already got that the slope was F'(x)=-2x
dan815
  • dan815
right
anonymous
  • anonymous
Right
dan815
  • dan815
now write the slope with those 2 points
ganeshie8
  • ganeshie8
Any line through \((5,0)\) is of form \[y=m(x-5)\tag{1}\] we want that line to "just touch" below parabola \[y=9-x^2\tag{2}\] simply set both equations equal to each other and use the discriminant to find the slope, \(m\)
dan815
  • dan815
and equate it to the F'(x)
dan815
  • dan815
|dw:1435217973229:dw|
anonymous
  • anonymous
Yeah lol i was typing the same xD
anonymous
  • anonymous
We get a quadratic equation when we solve for a... that means we'll get 2 values for a...
anonymous
  • anonymous
a^2 - 10a + 9 = 0
anonymous
  • anonymous
thats a 9 :P
dan815
  • dan815
yes
ganeshie8
  • ganeshie8
Yes, two tangent lines can be drawn to a parabola form an external point
anonymous
  • anonymous
lol xD ok so a=1 and x=9 ? Is that right?
anonymous
  • anonymous
i mean a* = 9
ganeshie8
  • ganeshie8
|dw:1435218241639:dw|
ganeshie8
  • ganeshie8
a=9 is right, but how did u figure ?
anonymous
  • anonymous
but neither of the 2 points i found for 'a' lie under the x-axis
anonymous
  • anonymous
I solved the quadratic a^2 -10a +9
ganeshie8
  • ganeshie8
okay i see you simply discarded a=1 without any reason
anonymous
  • anonymous
nope i wrote that as well lol ^^ check again :P
anonymous
  • anonymous
What happens next?
ganeshie8
  • ganeshie8
Okay just making sure :) plugin x=a=9 into the equation, you do get a point under the x axis
anonymous
  • anonymous
I do? Ohh thats right! xD
anonymous
  • anonymous
So that means I get 2 tangent lines?
ganeshie8
  • ganeshie8
plugin x=a=1, you get y = 9-x^2 = 9-1 = 8 that means the tangent line touches the parabola at (1, 8) which is above x axis so discard this
anonymous
  • anonymous
Ohh nice!!
anonymous
  • anonymous
I love discarding stuff in math xD haha
ganeshie8
  • ganeshie8
lol we're very much alike :P
dan815
  • dan815
i am not following exactly what u are doing here ganeshie y=m(x-5) y=9-x^2
dan815
  • dan815
are u then setting m=-2x to solve it
dan815
  • dan815
arent 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
  • ganeshie8
nope set both equations equal to each other and use discriminant : m(x-5) = 9-x^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
  • ganeshie8
solving that gives the slope of two tangent lines passing thru (5, 0)
dan815
  • dan815
oh i see! what you mean i didnt understand what discriminant meant lol
dan815
  • dan815
nice trick
anonymous
  • anonymous
neither did I o.o XD lol
Astrophysics
  • Astrophysics
Lol that had me confused to but I was too scared to ask, nice one.
ganeshie8
  • ganeshie8
thats okay, stick to dan's method as it is more calculussy
Astrophysics
  • Astrophysics
I feel like an idiot now xD
anonymous
  • anonymous
No I totally understood everything we did!! I just didn't get what you were explaining to dan at the end xP
anonymous
  • anonymous
@Astrophysics pshhhh in that case i'm more of an idiot that everyone! XD lol
dan815
  • dan815
basically 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
  • ganeshie8
reminds me of 3 idiots movie
anonymous
  • anonymous
Ahhh i see
dan815
  • dan815
|dw:1435219094311:dw|
anonymous
  • anonymous
Lmao XD truuueee :D so much humour the night before my exam :D lol
anonymous
  • anonymous
hmmm i see it now
ganeshie8
  • ganeshie8
just to conclude : x=a = 9 gives slope=-2x=-2*9 = -18 so the equation of desired tangent line is y = -18(x-5)
dan815
  • dan815
right, and u know how from the quadratic equation the 2 differnet solutions come from that +/- part in squareroot
dan815
  • dan815
|dw:1435219226919:dw|
anonymous
  • anonymous
Yep! Got it all :)
dan815
  • dan815
k good
anonymous
  • anonymous
Thanks so much!!! :)
dan815
  • dan815
yeah sure =]
Astrophysics
  • Astrophysics
Thank you teachers!
anonymous
  • anonymous
Yeshh :D
Astrophysics
  • Astrophysics
Are all these questions part of a practice exam?
anonymous
  • anonymous
Yeah, they're some sample/practice questions... or whatever they call it XD lol
anonymous
  • anonymous
@Astrophysics :D sorry about the late response, wasn't paying attention lol
Astrophysics
  • Astrophysics
Lol no worries was helping someone anyways, haha and good luck, calc exams have always been long >.<
anonymous
  • anonymous
thank you! :) Yeah it's gonna be 2.5 hours 8:30 in the morning >.<
Astrophysics
  • Astrophysics
Eek, haha get some sleep as well! Later :P
anonymous
  • anonymous
I'd better xD lol otherwise its a big fat 0 xD haha

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