anonymous
  • anonymous
I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION... Suppose that l be a line through the points (0,3)and (5,-2) and tangent to the graph of the function \[F(x)=\frac{ x^2-p }{ 3x+1 }\] where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F. A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y-3 = 0 D) 4x+6y-37 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I will post the answer exactly after one hour.
mathslover
  • mathslover
Hint: Never challenge OpenStudy
anonymous
  • anonymous
m= -2-3/5-0 = -1 the equation will be y-3= -1(x-0) x+y-3 =0

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TheViper
  • TheViper
Right @mathslover :)
hartnn
  • hartnn
so its not D
anonymous
  • anonymous
@hartnn that means???
hartnn
  • hartnn
that means answer can be a,b or c
anonymous
  • anonymous
yep.... but the answer hasn't yet finished... @hartnn
anonymous
  • anonymous
I will the slope of the line through the two points
hartnn
  • hartnn
p=-7 B)9x+9y+37=0
hartnn
  • hartnn
solve y=3-x=F(x)=.... u get two roots of x, equate them because the line is tangent and can have only one intersection point. from this i got p=-7. now answer can only be either A or B so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.
anonymous
  • anonymous
PLZ DONT POST THE SOLUTION
anonymous
  • anonymous
I am here to post the answer.
anonymous
  • anonymous
ok 5min
anonymous
  • anonymous
ok
hartnn
  • hartnn
whats wrong with my answer? and method ??
anonymous
  • anonymous
w/c is your ans?
hartnn
  • hartnn
B)
hartnn
  • hartnn
and p=-7
anonymous
  • anonymous
correct but the approach not.
hartnn
  • hartnn
didn't u get p=-7, the same way as i got ?
anonymous
  • anonymous
yape
hartnn
  • hartnn
then ?
anonymous
  • anonymous
you use the options to answer the question, isn't it?
anonymous
  • anonymous
Time is up @sauravshakya
anonymous
  • anonymous
OK.......
anonymous
  • anonymous
Here is the answer
anonymous
  • anonymous
First of all what you have to do is finding l so as to get p \[l:\frac{ y-3 }{ x-0 }=\frac{ -2-3 }{ 5-0 }\] \[l:y=-x+3\]thus \[F(x)=\frac{ x^2-p }{ 3x+1} =-x+3=y\]simplifying the above expression \[4x^{2}-8x-p-3=0\]Now since l is tangent to F(x), the quadratic equation above has only one solution. That means \[b^{2}-4ac=0\] \[(-8)^{2}-4(4)(-p-3)=0\]thus \[(-8)^{2}-4(4)(-p-3)=0\]\[p=-7\]Now we get F(x) to be \[F(x)=\frac{ x^2+7 }{ 3x+1 }\]Therefore lets solve F’(x)=-1 so as to get the coordinate of the unknown line. \[F'(x)=\frac{ 2x(3x+1)-3(x^{2}+7) }{ (3x+1)^2 }=-1\] \[3x^{2}+2x-21=-9x^{2}-6x-1\] \[12x^{2}+8x-20=0\]thus \[x=1 or x=\frac{ -5 }{ 3 }\]and \[F(1)=2 and F(\frac{ -5 }{ 3 })=\frac{ -22 }{ 9 }\]Thus we got this two points (1,2) and (-5/3,-22/9). The two possible lines therefore are \[\frac{ y-2 }{ x-1 }=-1{\rightarrow}y=-x+3\]w/c is the given line l. And \[\frac{ y-(-\frac{ 22 }{9 }) }{ x-(-\frac{ 5 }{ 3 })}=-1{\rightarrow}9y+9x+37=0\]w/c is the answer.
hartnn
  • hartnn
why f'(x) = -1 ?
anonymous
  • anonymous
b/c the slope at that point is -1
anonymous
  • anonymous
Because the line l and the required line are parallel
anonymous
  • anonymous
So, both of their slope is -1
hartnn
  • hartnn
okk.
hartnn
  • hartnn
it wasn't that difficult.
anonymous
  • anonymous
yes but need smart approach
anonymous
  • anonymous
YEP....... Actually I did a algebra mistake, as usual
hartnn
  • hartnn
lol! checking options(when given) is the smartest approach!! :P
anonymous
  • anonymous
agreed with @hartnn
anonymous
  • anonymous
that saves time in exam
anonymous
  • anonymous
For exam only
hartnn
  • hartnn
atleast i got the answer...
anonymous
  • anonymous
yape
anonymous
  • anonymous
ok thanks for your help....... closing

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