Question

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:

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

is m=-4?

Answer 7

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

Answer 8

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

Answer 9

4

Answer 10

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

Answer 11

thank you so much!

Answer 12

:)