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.

- anonymous

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- dan815

first you need to find the slope at this point

- anonymous

-2x ?

- dan815

that is the function of all the slopes wrt to x

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## More answers

- 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

the point where it touches f(x) which is below the x axis?

- dan815

take a look at your point

- dan815

(5, 0)

- dan815

whats the x value of the point

- anonymous

5

- dan815

right so we need to know f'(5) = ?

- anonymous

But that point doesn't even lie on the curve...

- dan815

we just want the slope there at x=5

- anonymous

so that would be -2(5) = -10, right?

- dan815

oh wait 5,0 isnt on this curve really?

- dan815

u are right lol xD

- anonymous

Yeah cause try plugging it in the function

- anonymous

XD

- anonymous

I think we should first find a point that apparently lies somewhere under the x-axis

- dan815

k lets graph and see what exactly is going on here

- dan815

|dw:1435217659239:dw|

- dan815

|dw:1435217751709:dw|

- dan815

okay so eventually this has to happen, they want us to find this x,y

- anonymous

Yep!

- anonymous

So are we solving for the secant line?

- anonymous

between (5,0) and (a, 9-a^2)

- dan815

you already got that the slope was
F'(x)=-2x

- dan815

right

- anonymous

Right

- dan815

now write the slope with those 2 points

- 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

and equate it to the F'(x)

- dan815

|dw:1435217973229:dw|

- anonymous

Yeah lol i was typing the same xD

- anonymous

We get a quadratic equation when we solve for a... that means we'll get 2 values for a...

- anonymous

a^2 - 10a + 9 = 0

- anonymous

thats a 9 :P

- dan815

yes

- ganeshie8

Yes, two tangent lines can be drawn to a parabola form an external point

- anonymous

lol xD ok so a=1 and x=9 ? Is that right?

- anonymous

i mean a* = 9

- ganeshie8

|dw:1435218241639:dw|

- ganeshie8

a=9 is right, but how did u figure ?

- anonymous

but neither of the 2 points i found for 'a' lie under the x-axis

- anonymous

I solved the quadratic a^2 -10a +9

- ganeshie8

okay i see you simply discarded a=1 without any reason

- anonymous

nope i wrote that as well lol ^^ check again :P

- anonymous

What happens next?

- ganeshie8

Okay just making sure :)
plugin x=a=9 into the equation, you do get a point under the x axis

- anonymous

I do? Ohh thats right! xD

- anonymous

So that means I get 2 tangent lines?

- 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

Ohh nice!!

- anonymous

I love discarding stuff in math xD haha

- ganeshie8

lol we're very much alike :P

- dan815

i am not following exactly what u are doing here ganeshie
y=m(x-5)
y=9-x^2

- dan815

are u then setting m=-2x to solve it

- 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

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

solving that gives the slope of two tangent lines passing thru (5, 0)

- dan815

oh i see! what you mean i didnt understand what discriminant meant lol

- dan815

nice trick

- anonymous

neither did I o.o XD lol

- Astrophysics

Lol that had me confused to but I was too scared to ask, nice one.

- ganeshie8

thats okay, stick to dan's method as it is more calculussy

- Astrophysics

I feel like an idiot now xD

- anonymous

No I totally understood everything we did!! I just didn't get what you were explaining to dan at the end xP

- anonymous

@Astrophysics pshhhh in that case i'm more of an idiot that everyone! XD lol

- 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

reminds me of 3 idiots movie

- anonymous

Ahhh i see

- dan815

|dw:1435219094311:dw|

- anonymous

Lmao XD truuueee :D so much humour the night before my exam :D lol

- anonymous

hmmm i see it now

- 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

right, and u know how from the quadratic equation the 2 differnet solutions come from that +/- part in squareroot

- dan815

|dw:1435219226919:dw|

- anonymous

Yep! Got it all :)

- dan815

k good

- anonymous

Thanks so much!!! :)

- dan815

yeah sure =]

- Astrophysics

Thank you teachers!

- anonymous

Yeshh :D

- Astrophysics

Are all these questions part of a practice exam?

- anonymous

Yeah, they're some sample/practice questions... or whatever they call it XD lol

- anonymous

@Astrophysics :D sorry about the late response, wasn't paying attention lol

- Astrophysics

Lol no worries was helping someone anyways, haha and good luck, calc exams have always been long >.<

- anonymous

thank you! :) Yeah it's gonna be 2.5 hours 8:30 in the morning >.<

- Astrophysics

Eek, haha get some sleep as well! Later :P

- anonymous

I'd better xD lol otherwise its a big fat 0 xD haha

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