## 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:

1. Michele_Laino

the equation of the tangent line, with slope m and which passes at the point (0,-2), is: $\Large y - \left( { - 2} \right) = m\left( {x - 0} \right)$

2. amy0799

I understand that, but how do I get the positive and negative slope?

3. 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

4. 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)

5. 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

6. amy0799

is m=-4?

7. Michele_Laino

yes! nevertheless it is one solution, we have to find the second solution

8. Michele_Laino

$\Large {m^2} - 16 = 0$ what is m?

9. amy0799

4

10. Michele_Laino

correct! your tangent lines are: y=4x-2 y=-4x-2

11. amy0799

thank you so much!

12. Michele_Laino

:)