Question 1188448
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Determine the restrictions on the y-intercept so that y = 3x^2 + 6x - 1 intersects 
with a line with slope 2 in more than one place.
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        The condition in the post is over-twisted, 

        so I will re-formulate it to make its meaning clear.



<pre>
            Determine the restrictions on the y-intercept of the straight line of the slope 2

            so it intersect the curve y = 3x^2 + 6x - 1 in more than one point.
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>SOLUTION</U>



<pre>
The straight line equation is  y = 2x + k,  where "k" is the y-interception value.

We should determine this value of "k" from the condition that the equation

    3x^2 + 6x - 1 = 2x + k


has more than one solution.


We rewrite this equation in standard quadratic form

    3x^2 + 4x - (k+1) = 0.


The condition that it has more than one real solution means that the discriminant is positive

    d = b^2 - 4ac > 0,  where  a= 3;  b= 4  and  c= -(k+1).


So, the discriminant is  

    d = 16 + 4*3*(k+1) = 16 + 12(k+1).


The condition d > o is

    16 + 12(k+1) > 0

or

    12(k+1) > -16

       k+1 > - {{{16/12}}} = - {{{4/3}}}

         k > - {{{4/3}}} - 1 = - {{{7/3}}}.


<U>ANSWER</U>.  The y-intercept "k" must be greater than  - {{{7/3}}} = -2 {{{1/3}}} :  k > -2 {{{1/3}}}.
</pre>

Solved.