Question 291563
Take the first derivative of your parabola function.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'\ =\ -6x\ -\ 1]


The slope of the line is 4, so set the derivative function equal to that slope:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6x\ -\ 1\ =\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{5}{6}]


If the line were actually tangent, then the point of tangency would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{5}{6},f\left(-\frac{5}{6}\right)\right)]


Do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3\left(-\frac{5}{6}\right)^2\ -\ \left(-\frac{5}{6}\right)\ +\ 4\ =\ \frac{11}{4}]


So, for a tangent line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{11}{4}\ =\ 4\left(-\frac{5}{6}\right)\ +\ k]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \frac{73}{12}]


So, if the line must not intersect the parabola:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ >\ \frac{73}{12}]