For how many integer values of $m$ parabola $y=(m-2)x^2+12x+m+3$ passes through only three quadrants?

Question

For how many integer values of $m$ the graph of parabola $y=(m-2)x^2+12x+m+3$ passes through only three quadrants?

$1)0\qquad\qquad2)7\qquad\qquad3)12\qquad\qquad4)\text{infinity}$

If the parabola $ax^2+bx+c=0$ passes through three quadrant I think we should have $\frac ca>0$ and $\Delta>0$:

$$\frac{m+3}{m-2}>0\Rightarrow m\in(-\infty,-3] \cup(2,+\infty)$$

$$\Delta'>0\Rightarrow 36-(m-2)(m+3)>0\Rightarrow m^2+m-42<0\Rightarrow m\in [-7,6]$$

Also for $m=2$ we have $y=12x+5$ and it passes through three quadrants.

So $m$ can be $-7,-6,\cdots,-3, $ or $2,3,4,5,6$ . so there are $10$ possible values for $m$ but this isn't in the options. What am I missing?

Answer 1

One mistake is that

$$ m^2+m-42<0$$

gives $m\in (-7,6)$ instead of $m\in [-7, 6]$. So $\{-7, 6\}$ are excluded.

If one excludes also $m=2$ (which does not give a parabola), then there are only $7$ possible choices: $\{-6, -5, -4, -3, 3, 4, 5\}$.

Answer 2

I was just doing the same observation as @Arctic Char.

Here is a graphical representation confirming that (the values of $m$ are superimposed on the curves):