amy0799
  • amy0799
Find equations of the two tangent lines to the graph of f(x) = 2x^2 that pass through the point (0, -2). line with positive slope: line with negative slope:
Mathematics
schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
the equation of the tangent line, with slope m and which passes at the point (0,-2), is: \[\Large y - \left( { - 2} \right) = m\left( {x - 0} \right)\]
amy0799
  • amy0799
I understand that, but how do I get the positive and negative slope?
Michele_Laino
  • Michele_Laino
I simplify that equation like below: \[\Large y = mx - 2\] then I consider this algebraic system: \[\Large \left\{ \begin{gathered} y = mx - 2 \hfill \\ y = 2{x^2} \hfill \\ \end{gathered} \right.\] I solve it, substitution method, namely I replace y into the second equation, by mx-2, so I get this: \[\Large mx - 2 = 2{x^2}\] now, in order to get the requested values for m, we have to request that the discriminant of that equation is equal to zero

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amy0799
  • amy0799
i found the derivative to be 4x. and plugging in 0 to 4x gets me o. so would the equations be: y+2=0(x-0) y+2=-0(x-0)
Michele_Laino
  • Michele_Laino
I rewrite my last equation as below: \[\Large 2{x^2} - mx + 2 = 0\] which is a quadratic equation. Now I request that its discriminant, \Delta is equal to zero. \[\Large \Delta = {b^2} - 4ac = {\left( { - m} \right)^2} - 4 \cdot 2 \cdot 2 = 0\] Please, solve for m
amy0799
  • amy0799
is m=-4?
Michele_Laino
  • Michele_Laino
yes! nevertheless it is one solution, we have to find the second solution
Michele_Laino
  • Michele_Laino
\[\Large {m^2} - 16 = 0\] what is m?
amy0799
  • amy0799
4
Michele_Laino
  • Michele_Laino
correct! your tangent lines are: y=4x-2 y=-4x-2
amy0799
  • amy0799
thank you so much!
Michele_Laino
  • Michele_Laino
:)

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