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.

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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
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first you need to find the slope at this point
-2x ?
that is the function of all the slopes wrt to x

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Other answers:

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

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