The graph of the equation y=ax^2+bx-6 is completely below the x axis. If , a^2=49 what is the largest possible integral value of b?
This describes an upside down parabola (opens downward...looks like a dome)
so a^2 = 49 means a = -7
so you have -7x^2+bx-6 =0 Using quadratic formula
\(0> {-b \pm \sqrt{b^2-4(7)(-6)} \over -14}\) yields b <12.96 and b > -12.96
so b<12.96 is the greatest value