(1) It is clear that all parabolas pass through $(2,9)$, and the vertices lie on the parabola $y = -x^2+4x+5$. (2) The larger root is $ \frac{-a-2+\sqrt{a^2+12a}}{2}$, so $a \in (-\infty , -12)$ or $(0,\infty )$. Now, as $a$ gets smaller, the root gets bigger. On the other hand, the root is negative in $(0,\frac{1}{2})$, and the function is concave, so the lower bound is $-1$.

note that , irrespective of the value of a , the parabola passes through $(2,9)$ note that , the vertex is $(-\frac{a+2}{2},1-2a-[\frac{a+2}{2}]^2)$ and these points lie on the parabola $y=5+4x-x^2$. now , we come to part (b) note that for having 2 distinct real roots , we must have $D=a(a+12)>0$ , i.e. $a>0$ , or $a<-12$ now , note that , the larger root is $\frac{-(a+2)+\sqrt{a^2+12a}}{2}$ if $a>0$ , then the range of this root is clearly $(-1,\infty)$ . [for finding this , we calculate the derivative , check its sign for a>0 ,find limits etc. (everything for a>0)] we can similarly find it for $a<-12$ using simple calculus.

For $a<-12$, it is clear that the range is (5,infinity), but then for $a>0$, isn't it that the range is only $(-1,2)$ by using limits and derivatives?

Cause like for the $a>0$ case we have $a<\sqrt{a^2+12a}<a+6$, so it's clearly bounded between $(-1,2)$