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amy0799

  • one year ago

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:

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  1. Michele_Laino
    • one year ago
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    the equation of the tangent line, with slope m and which passes at the point (0,-2), is: \[\Large y - \left( { - 2} \right) = m\left( {x - 0} \right)\]

  2. amy0799
    • one year ago
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    I understand that, but how do I get the positive and negative slope?

  3. Michele_Laino
    • one year ago
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    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

  4. amy0799
    • one year ago
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    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)

  5. Michele_Laino
    • one year ago
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    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

  6. amy0799
    • one year ago
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    is m=-4?

  7. Michele_Laino
    • one year ago
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    yes! nevertheless it is one solution, we have to find the second solution

  8. Michele_Laino
    • one year ago
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    \[\Large {m^2} - 16 = 0\] what is m?

  9. amy0799
    • one year ago
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    4

  10. Michele_Laino
    • one year ago
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    correct! your tangent lines are: y=4x-2 y=-4x-2

  11. amy0799
    • one year ago
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    thank you so much!

  12. Michele_Laino
    • one year ago
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    :)

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