Find the smallest $k\in \mathbb{Z}$ such that $f(x)=x^2-3x+k$ and $g(x)=x-2$ do not intersect.

$$\large{ \text{Here are some instructions from the original question: } \ \\ }$$

$$\large{ k \in \mathbb{Z} \ \ \land \\ f(x)=x^{2}-3x+k \ \ \land \ \ g(x)=x-2 \ \\ \text{And, there is NO any intersection point between f(x) and g(x). } \ \\ k_{(min) \, }= \;? } \ \\$$

$$\large{ \text{Here is just an example graph which I've drawn after seeing the answer: } \ \\ }$$

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):

$$\large{ (-4)^{2}- \Big( 4 \times (1) \times (k+2) \Big) = \Delta \ \\ 4^{2}- \Big( 4 (k+2) \Big)= 16-(4k+8)= \Delta \ \\ \Delta \lt{0} \iff 16-4k-8 \lt{0} \iff 8-4k \lt{0} \ \\ }$$ $$\text{Here comes a simple method to reach the answer: } \ \\ \large{ 8-4k=0 \ \\ 4k=8 \ \\ k=2 } \ \\ \text{but,} \ \ k_{min} \ \text{ equals just one of these } \ \mathbb{R^{+}} \ \text{and} \ \mathbb{Z^{+}} \ \text{numbers} : \ (2, \, {+}\infty ) \ \\ \text{So, } \ \large{ k_{min}=3 }$$

-

You want the quadratic, $\Delta$, to have no solution. That means you want its discriminant, which is $4^2-4(k+2)$, to be negative. In particular, you want the smallest integer $k$ that makes it negative. Can you take it from there?

-
Thanks for the help, Gerry. But, where that $\, \large{ 4^{2}-4(k+2) \, }$ comes from, I still can't see it...? Of course, I could calculate that, I mean, if I could able to see and get it... :) And, my mechanical pencil and papers're dying for that... :D :) – Kerim Atasoy Apr 30 '12 at 0:39
@Kerim: if you still remember the quadratic formula, the discriminant is the stuff within the square root in that formula. – J. M. Apr 30 '12 at 1:06
@J.M. : Thanks for the help. I may be see it later... – Kerim Atasoy Apr 30 '12 at 1:18
@KerimAtasoy: It is similar to completing the square. You want to write $\Delta=(x \pm something)^2 \pm something else$ The first term is always $\ge 0$. The second will involve $k$ and if it is $\gt 0$, then you know $\Delta \gt 0$ and there is no intersection. There will not be a minimum $k$ to have no intersection; there will be a maximum $k$ to have one (which will be one point only, not two). – Ross Millikan Apr 30 '12 at 2:28
@RossMillikan: Well, I think, I finally understand my problem, please, let me know if I did it wrong or right... :) – Kerim Atasoy Apr 30 '12 at 15:01

Hint: You want $\Delta \gt 0$ as it measures the distance $f$ is above $g$ at any given $x$. If $k$ is large enough, $\Delta$ will be a perfect square plus something and squares cannot be negative.

-
Thanks for the hint, Ross, I'm still thinking over it... – Kerim Atasoy Apr 30 '12 at 0:30

Since it seems like a Homework question i'd rather give a hint. You got the distance between the graphs of f and g as: $$d(f,x)=f(x)-g(x)=x^2-3x+k-(x-2)=x^2-4x+2+k$$This must always be POSITIVE and not negative since $G_f$ is above $G_g$ for all $x\in R$.Therefore the equation shouldn't have any real roots, in other words the discriminant should be negative: $$\Delta<0$$Watch out while calculating the discriminant. It shouldn't contain any x's. By solving the inequality you will find the minimum integer k that satisfies it.

-
Actually, it's not a homework question... Thanks for the help...:) – Kerim Atasoy Apr 30 '12 at 0:42