Finding point of contact and value of b If a line $y = 2x - b$ connects with a curve $y = 3x^2 + 2$, how do I approach finding the value of b? To find the coordinates, if I'm not mistaken, I need to set the derivative of $y = 3x^2 + 2$ equal to the equation of the line. Is this correct? Many thanks in advance!
 A: Your "curve", a parabola, takes its minimum value of $2$ at $x = 0$.  Imagine $b$ to be very large; your line will miss your curve.  As $b$ decreases, the line moves upward, always parallel to itself.  When it first touches the parabola, it will be tangent at the point of conract.  This can be rigorously proved, but I'm simply asking you to use your geometric intuition.  Since it's tangent, the slopes of the two "curves" are the same.  The line has constant slope $2$, the parabola slope $6x$.  $6x = 2$ implies $x = \frac{1}{3}$.  The $y$-value of the parabola there is $\frac{7}{3}$.  Plugging $x$ and $y$ into the line equation, you can solve for $b$.  It is $-\frac{5}{3}$.
Hope this helped!  Cheers!
A: By differentiation, we have
$$\dfrac{dy}{dx} = 6x$$
Since we know that the slope of a line is $2$,
$$\dfrac{dy}{dx} = 2 = 6x$$
$$x = \frac{1}{3}$$
Therefore,
$$y = 3(\frac{1}{3})^2 + 2$$
$$y = \frac{7}{3}$$
Thus,
$$y - \frac{7}{3} = 2(x - \frac{1}{3})$$
$$y = 2x + \frac{5}{3}$$
So $b = \frac{5}{3}$.
