it is given the family of quadratic funcion f(x)=2x2+(m+6)x+8. For which m-s will the graph will lie above the alphabetical axis

To determine for which values of "m" the graph of the quadratic function $$\(f(x) = 2x^2 + (m+6)x + 8\)$$ will lie above the x-axis (i.e., be entirely positive), you need to consider the discriminant of the quadratic equation. The discriminant D of a quadratic equation in the form $$\(ax^2 + bx + c\)$$ is given by: $$\[D = b^2 - 4ac\]$$ In your case, (a = 2), (b = m+6), and (c = 8). To ensure that the graph is above the x-axis, you want (D) to be negative because a negative discriminant means that the quadratic equation has no real roots (i.e., it doesn't intersect the x-axis). So, you need: $$\[D = (m+6)^2 - 4(2)(8) < 0\]$$ Now, you can solve for "m" to find the range of values: $$\[(m+6)^2 - 64 < 0\]$$ Take the square root of both sides: $$\[|m+6| < 8\]$$ This means that: $$\[-8 < m+6 < 8\]$$ Now, subtract 6 from all sides: $$\[-14 < m < 2\]$$ So, the graph of the quadratic function $$\(f(x) = 2x^2 + (m+6)x + 8\)$$ will lie above the x-axis when (m) is in the open interval (-14, 2).