Condition for two functions to never intersect Given $y=mx+b$ and $y=|x+1|$ and $-1\leq m\leq1$ find the relationship between $b$ and $m$ so that the two graphs never intersect.
Is there a purely algebraical way to logically derive a solution or is it easier to inspect a graph and deduce it? 
 A: Tinkering with the Desmos graph here we note that $|m| > 1$ will lead to a line with a slope so steep that, eventually, it intersects the absolute value function $y = |x+1|$, which appears as a graph with vertex at $(-1, 0)$ and opens upwards with rays of slope $\pm 1$.
In the case of $m=1$, the only way to avoid intersection is to have $b<1$; in the case of $m=-1$, the only way to avoid intersection is to have $b<-1$. To see why, experiment with the linked graph. Otherwise, we have that $|m| < 1$, and so we look at the two cases of $0 \leq m < 1$ and $-1 < m < 0$.
In either case, we have a line whose slope is less steep than the (absolute value of the) rays' slopes; so, we simply need to ensure that the linear function is below the absolute value function. To see when this happens, it suffices to find the $b$ for which the line intersects the vertex at $(-1,0)$; we can then use any lower value for $b$ to ensure that the line is below the absolute value.
As to an algebraic solution: For the line $y=mx+b$ to be satisfied by the point $(-1,0)$ means that the equation $0 = m(-1) + b$ is true; i.e., that $b = m$. As described above, it now suffices to pick any smaller value of $b$; so, the functions' graphs never intersect provided $b < m$.
To see this in Desmos, here are the graphs where $b = m - a$ for $a > 0$; the parameters are set to play, and there is also depicted the graph of $y = |x+1|-(mx+b)$, which would have an output of $0$ if, and only if, your two original functions intersected. As you can see, that graph's $y$-values are all strictly greater than $0$.
A: I tried squaring both sides to get rid of the absolute value but it ended up with a function that didn't help for some reason. I'll pursue that further afterwards.
$$x≥-1$$
$$mx+b=x+1$$
$$x= \frac{1-b}{m-1}$$
$$−1≤m≤1 \\ −2≤m-1≤0$$
$$solution \ doesn't \ exist\  if \ x<-1$$
$$\frac{1-b}{m-1}<-1$$
$$1-b>-m+1$$
$$m>b$$
$$x≤-1 \ yields \ same \ result$$
