$ \large{ \text{Here are some instructions from the original question: } \ \\ } $
$ \large{ k \in \mathbb{Z} \ \ \land \\ f(x)=x^{2}-3x+k \ \ \land \ \ g(x)=x-2 \ \\ \text{And, there is NO any intersection point between $f(x)$ and $g(x)$. } \ \\ k_{(min) \, }= \;? } \ \\ $
$ \large{ \text{Here is just an example graph which I've drawn after seeing the answer: } \ \\ } $
I've tried this so far:
$ \large{ f(x) \ne g(x) \; \land \; f(x)-g(x)\lt{0} \leftrightarrow \Delta \lt{0} \; \\ \Delta=x^{2}-4x+k+2 \; ... \\ } $
The answer lies just behind that $\Delta$ but, unfortunately, I couldn't able to further the solution anymore...and, I need your help from here, so, please, show me how to do that.
Thank you very much...
PS: Finally, I understand it now... :) Thank you so much for your all help!...
A very big hint: $ \large{ \Delta= \Big(b \Big)^{2}- \Bigg(4 \times \Big( a \Big) \times \Big( c \Big) \Bigg) \ (1) \\ } $
Now, I'm furthering my solution using that* equation *(1):
$ \large{ (-4)^{2}- \Big( 4 \times (1) \times (k+2) \Big) = \Delta \ \\ 4^{2}- \Big( 4 (k+2) \Big)= 16-(4k+8)= \Delta \ \\ \Delta \lt{0} \iff 16-4k-8 \lt{0} \iff 8-4k \lt{0} \ \\ } $ $ \text{Here comes a simple method to reach the answer: } \ \\ \large{ 8-4k=0 \ \\ 4k=8 \ \\ k=2 } \ \\ \text{but,} \ \ k_{min} \ \text{ equals just one of these } \ \mathbb{R^{+}} \ \text{and} \ \mathbb{Z^{+}} \ \text{numbers} : \ (2, \, {+}\infty ) \ \\ \text{So, } \ \large{ k_{min}=3 } $