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

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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

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I will post the answer exactly after one hour.
Hint: Never challenge OpenStudy
m= -2-3/5-0 = -1 the equation will be y-3= -1(x-0) x+y-3 =0

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Other answers:

Right @mathslover :)
so its not D
@hartnn that means???
that means answer can be a,b or c
yep.... but the answer hasn't yet finished... @hartnn
I will the slope of the line through the two points
p=-7 B)9x+9y+37=0
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.
PLZ DONT POST THE SOLUTION
I am here to post the answer.
ok 5min
ok
whats wrong with my answer? and method ??
w/c is your ans?
B)
and p=-7
correct but the approach not.
didn't u get p=-7, the same way as i got ?
yape
then ?
you use the options to answer the question, isn't it?
Time is up @sauravshakya
OK.......
Here is the answer
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.
why f'(x) = -1 ?
b/c the slope at that point is -1
Because the line l and the required line are parallel
So, both of their slope is -1
okk.
it wasn't that difficult.
yes but need smart approach
YEP....... Actually I did a algebra mistake, as usual
lol! checking options(when given) is the smartest approach!! :P
agreed with @hartnn
that saves time in exam
For exam only
atleast i got the answer...
yape
ok thanks for your help....... closing

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