First equation represents a parabola on a coordinate plane, opened up.
Second equation represents a straight line with the slope of 2 unit.


The following cases are possible:


    (a) straight line intersects the parabola in two points.

    (b) straight line touches the parabola (having only one common point with the parabola).

    (c) straight line is located below the parabola and does not have intersection points with the parabola.


To analyze these cases, we consider the system of equations

    y = x*(x-8)    (1)
    y = 2x + k     (2)


From equations (1) and (2), we have this equation

    x*(x-8) = 2x +k.


Simplify it

    x^2 - 8x = 2x + k,

    x^2 - 10x - k = 0.


The last equation is a standard form quadratic equation.
All the cases are controlled by the discriminant of this quadratic equation.

Referring to the general form quadratic equation, the discriminant is

    d = b^2 - 4ac.


In your problem,

    d = (-10)^2 - 4*1*(-k) = 100 + 4k.



Case (a) (two solutions and two intersection points) may happen if and only if the discriminant is positive

    100+4k > 0,  which implies  k > -100/4 = -25.



Case (b) (one solution, touching) may happen if and only if the discriminant is zero

    100+4k = 0,  which implies  k = -100/4 = -25.



Case (c) (no solutions, no intersection points) may happen if and only if the discriminant is negative

    100+4k < 0,  which implies  k < -100/4 = -25.


Thus, you have this ANSWER

    - there are two intersection point if and only if k > -25;

    - there is only one common point (touching) if and only if k = -25;

    - there are no intersection points if and only if k < -25.


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and use free of charge plotting tool there.


Print your given equations there, taking appropriate numerical value of k (k= -30;  k= -25;  k= -20)
and obtain the required plots momentarily.