SOLUTION: determine k such that g(x)=3x^2+8x+k intersects f(x)=2x-1 at ine point.

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Question 1162180: determine k such that g(x)=3x^2+8x+k intersects f(x)=2x-1 at ine point.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
find a place where the line 2x-1 is tangent to the curve.
Want one point only, where the slopes are the same, 2, because that is the slope of the line
find slope of the quadratic: derivative is 6x+8, and that equals 2 when x=-1
when x=-1, 2x-1=-3
so the point is (-1, -3) for the line. It has to be for the quadratic as well.
when x=-1, the quadratic is 3-8+k and that has to equal -3
-5+k=-3
k=2

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

They want you find the value of "k" in a way that the straight line f(x) = 2x-1 be a tangent line 
to the parabola  g(x) = 3x^2 + 8x + k.


It means that the equation

    3x^2 + 8x + k = 2x - 1 

has only one real root (the two roots merge into one root).


The equation is equivalent to

    3x^2 + 6x + (k+1) = 0.     (*)


Its discriminant is


    b^2 - 4ac =  = 36 - 12(k+1)


We want the discriminant be zero

    36 - 12(k+1) = 0,   or   36 = 12(k+1),   k+1 = 36/12 = 3,  k = 3-1 = 2.


ANSWER.  k = 2.


CHECK.  At k= 2, the equation (*)  becomes  3x^3 +6x + 3 = 0.

        It is equivalent to  x^2 + 2x + 1 = 0,  which  factors into   = 0  and has only one real solution x= -1.


        The point  (x,f(x)) = (-1,f(-1)) = (-1,-3)  is the same as  the point  (x,g(x)) = (-1,g(-1)) = (-1,-3).


        It is the unique common point of the two given lines.

Solved.   //   The problem was solved using  Algebra only,  without using  Calculus.



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