How to prove a line is above another line Suppose I have the following line:
$y=-4x + 80$, for $x \ge 0$ and $y \ge 0$
I want to show that if I vary the slope, $m$ like so:
$-4\lt m \le -2$
Then the new line will be above the old line (except when $x=0$).
How do I formally show this?
My attempt:

With the new slopes, $y$ can be:
$-4x+80 \lt y \le -2x+80$
Subtracting the old line ($y=-4x+80$) yields:
$0 \lt y \le 2x$
Which says the difference between the old and new lines can range from $0$ upto and including $2x$.

Is it sufficient or is there a  more convincing way to prove this?
 A: If you want to make so detailed an argument, I’d make it more like this.
You’re starting with the line $y=f(x)$, where $f(x)=-4x+80$. Now let $y=g(x)$, where $g(x)=mx+80$, be any straight line also passing through $\langle 0,80\rangle$. For any $x\in\Bbb R$ the second line lies above the first at $x$ if and only if $g(x)>f(x)$, or $g(x)-f(x)>0$. Now
$$g(x)-f(x)=(mx+80)-(-4x+80)=(m+4)x\;,$$
and certainly $(m+4)x>0$ whenever $m+4>0$ and $x>0$. Since you’re looking only at values of $m$ greater than $-4$, you can conclude that the second line lies above the first for all $x>0$.
This directly addresses the relationship between the heights of the two lines at a particular value of $x$; your approach silently sneaks in the fact that $g(x)$ increases with $m$ without justifying it.
A: More convincing could be to find an expression for the intersection of the old line with the new in terms of m, and show it's negative. Then deduce that because at x = 1 the new one is above the old one, it's above the old one for all x bigger than the intersection (which includes all positive x). 
I feel like it's convincing enough to just say "they agree at 0, and the new line's y value decreases slower" though.
