nenadmatematika
  • nenadmatematika
Choose c so that y=5x-7 is tangent to y=x^2+cx
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
you havent said at what point on the second graph, c could be loads of different values
nenadmatematika
  • nenadmatematika
oh, thank you for that!! :D that's what I was thinking, it seemed to me like incomplete task :D
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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
nenadmatematika
  • nenadmatematika
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
anonymous
  • anonymous
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 + (c-5)x + 7
nenadmatematika
  • nenadmatematika
I tried and I get some ugly values for c......
anonymous
  • anonymous
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
nenadmatematika
  • nenadmatematika
that's right I'm sure that's the problem!!! thank you for your effort :D
anonymous
  • anonymous
you're welcome :)
asnaseer
  • asnaseer
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 \(y=5x-7\)) so we get:\[5=2x+c\]\[\therefore x=\frac{5-c}{2}\tag{a}\]Since the line \(y=5x-7\) is supposed to be a tangent to the curve \(y=x^2+cx\), then it must touch the curve at some point, therefore:\[5x-7=x^2+cx\tag{b}\]Substituting (a) into (b) we get:\[5*\frac{5-c}{2}-7=(\frac{5-c}{2})^2+c*\frac{5-c}{2}=\frac{(5-c)^2}{4}+\frac{c(5-c)}{2}\]multiplying both sides by 4 gives:\[10(5-c)-28=(5-c)^2+2c(5-c)\]\[50-10c-28=25-10c+c^2+10c-2c^2\]\[c^2-10c-3=0\]Solve this to get the two possible values for c.
nenadmatematika
  • nenadmatematika
that's what I got and when I saw sqrt(112) I knew that something is wrong :D thank you!
asnaseer
  • asnaseer
yw
anonymous
  • anonymous
find the slope of each line i.e y=5x-7 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=-1/5-2x

Looking for something else?

Not the answer you are looking for? Search for more explanations.