### Instruction: 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 $b$?### Response: I notice that the graph of the equation is a parabola that opens either up or down depending on the sign of $a$.
Since the graph is completely below the $x$-axis, that means the parabola has no real roots, or equivalently, the discriminant of the quadratic equation is negative.
The discriminant is $b^2-4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.
In this case, $c=-6$, so the discriminant is $b^2-4a(-6)$.
I want this to be negative, so I need to have $b^2<24a$.
I also know that $a^2=49$, so $a=\pm 7$.
If $a=7$, then $b^2<24(7)$, or $b^2<168$.
The largest integer that satisfies this inequality is $b=12$.
If $a=-7$, then $b^2<24(-7)$, or $b^2<-168$.
This is impossible, since $b^2$ is always nonnegative.
Therefore, the only possible value of $a$ is $a=7$, and the largest possible integral value of $b$ is $b=12$.