For what values of $\;$ $a$ $\;$ do the graphs of the functions $\;$ $y = 2ax + 1$ $\;$ and $\;$ $y = \left(a - 6\right) x^2 - 2$ $\;$ do not intersect?
The given functions are: $\;$ $y = 2ax + 1$ $\;\;\; \cdots \; (1a)$ $\;$ and
$y = \left(a - 6\right) x^2 - 2$ $\;\;\; \cdots \; (1b)$
When functions $(1a)$ and $(1b)$ intersect, we have,
$2ax + 1 = \left(a - 6\right) x^2 - 2$
i.e. $\;$ $\left(a - 6\right) x^2 - 2ax - 3 = 0$ $\;\;\; \cdots \; (2)$
Now, $(1a)$ and $(1b)$ will not intersect when equation $(2)$ does not have any real roots
i.e. $\;$ when discriminant $\Delta$ of $(2)$ is less than $0$.
i.e. $\;$ Discriminant $\Delta = \left(-2a\right)^2 - 4 \times \left(a - 6\right) \times \left(-3\right) < 0$
i.e. $\;$ $4a^2 + 12a - 72< 0$
i.e. $\;$ $a^2 + 3a - 18 < 0$
i.e. $\;$ $\left(a - 3\right) \left(a + 6\right) < 0$
i.e. $\;$ $a < 3$ $\;$ and $\;$ $a > -6$ $\;\;\;$ OR $\;\;\;$ $a > 3$ $\;$ and $\;$ $a < -6$
i.e. $\;$ $-6 < a < 3$ $\;\;\;$ OR $\;\;\;$ Condition $a > 3 \cap a < -6 = \phi$ (null set)
$\therefore \;$ Curves $(1a)$ and $(1b)$ will not intersect when $a \in \left(-6, 3\right)$.