1. restrictions

determine the restrictions on the y-intercept so that $y= 3x^2+6x-1$ intersects with a line with slope 2 in more then one place.

2. restrictions

I'll take a shot at this...

1. Differentiate the equation to find the x,y coordinates where the slope=2: The derivative is 6x+6, which is equal to 2 where X = -2/3. Plugging -2/3 into your original equation gives us Y = -11/3.

2. If we want the line Y=2x+b to be tangential to this (i.e., touch it only once), we'd need to solve for b using the above x,y pairing:
-11/3 = 2* (-2/3) + b. If I did the math correctly, b = -7/3.

3. So if the y intercept is -7/3, we intersect your equation at only one point. If the y intercept is less than -7/3 (i.e., more negative), we wouldn't intersect your equation at all. So I believe the answer is that as long as the b in Y=2x+b is strictly greater than -7/3, we ensure that we cross through the line your equation represents exactly twice.

- Steve J

3. Hello, euclid2!

I have a different approach. .(Is anyone surprised?)

Determine the restrictions on the y-intercept so that $y\:=\: 3x^2+6x-1$
intersects a line with slope 2 in more then one place.
The equation of the line is: . $y \:=\:2x+b$

Find the intersections of the parabola and the line:

. . $3x^2 + 6x - 1 \:=\:2x + b \quad\Rightarrow\quad 3x^2 + 4x - (b+1) \:=\:0$

Quadratic Formula: . $x \;=\;\frac{-4 \pm\sqrt{4^2 - (4)(3)[-(b+1)]}}{2(3)} \;=\;\frac{-4 \pm\sqrt{28 + 12b}}{6}$

There will be two roots (two intersections) if the discriminant is positive.

. . Therefore: . $28 + 12b \:>\:0 \quad\Rightarrow\quad b \:>\:-\frac{7}{3}$

We agree, Steve_J !
.