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08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
you havent said at what point on the second graph, c could be loads of different values
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
oh, thank you for that!! :D that's what I was thinking, it seemed to me like incomplete task :D
 2 years ago

08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
if they did give you the point, you would firstly differentiate y, then set y' = 5 and solve for c given the value of x in the question
 2 years ago

08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
actually, you could also solve for c by taking eq1 from eq2, because you would get a quadratic, that is probably why they didnt give a point
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
thank you very much, I'm actually good at this...My younger brother asked me to help him about this, and I got stucked.........Surely he didn't send me complete task :D
 2 years ago

08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
actually, this may be the whole task, as you can solve for where the two graphs cross (as at the tangent point the two graphs are equal. so you have to solve 0 = x^2 + (c5)x + 7
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
I tried and I get some ugly values for c......
 2 years ago

08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
I'm just guessing at what the question wants. seeing how complicated the question is getting, it is likely that the question is just missing the information on the actual tangent point
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
that's right I'm sure that's the problem!!! thank you for your effort :D
 2 years ago

08bkrishna Group TitleBest ResponseYou've already chosen the best response.3
you're welcome :)
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
the slope of the tangent to \(y=x^2+cx\) at any point would be given by its derivative:\[y'=2x+c\]We want the slope to equal 5 (since its equation must be \(y=5x7\)) so we get:\[5=2x+c\]\[\therefore x=\frac{5c}{2}\tag{a}\]Since the line \(y=5x7\) is supposed to be a tangent to the curve \(y=x^2+cx\), then it must touch the curve at some point, therefore:\[5x7=x^2+cx\tag{b}\]Substituting (a) into (b) we get:\[5*\frac{5c}{2}7=(\frac{5c}{2})^2+c*\frac{5c}{2}=\frac{(5c)^2}{4}+\frac{c(5c)}{2}\]multiplying both sides by 4 gives:\[10(5c)28=(5c)^2+2c(5c)\]\[5010c28=2510c+c^2+10c2c^2\]\[c^210c3=0\]Solve this to get the two possible values for c.
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
that's what I got and when I saw sqrt(112) I knew that something is wrong :D thank you!
 2 years ago

niki Group TitleBest ResponseYou've already chosen the best response.0
find the slope of each line i.e y=5x7 and y=x^2+cx slope for the first one is 5(differentiate wrt x) and the other is (2x+c).the 2 lines are perpendicular hence (2x+c).5=1 hence c=1/52x
 2 years ago
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