Modulus function: finding set of values. I didn't learn this in my syllabus but it is in some past papers. I need a small explanation for how to find the possible values for p to give 0, 1 or 2 roots.
Here is an example question: $f(x)=6-|x+2|$ with roots (-8,0) and (4,0)
Find the set of values for p for which the equation $f(x)=px+5$ has 0 solutions or 1 solution or 2 solutions.
I tried to equate both versions of the modulus to the second definition of the function but it didn't really get me anywhere.
 A: I think what you mean it to find the values for which $6-|x+2| = px + 5$ have $0, 1$ or $2$ solutions.  The solutions, not roots, are when $6-|x+2|$ and $px+5$ both equal the same number which might not and probably will not be $0$.  $6-|x+2|$ has two roots; $x=4$ and $x = -8$ and $px +5$ will have one root if $p\ne 0$ and that is $x =-\frac 5p$.  But we are not looking for when they are both $0$ (which can only happen if $-\frac 5p =-8$ or if $-\frac 5p = 4$) but when they are both equal to .... something.
If $6-|x+2| = px + 5$ then we have two possible cases and two equations to solve.

*

*$x+2 \ge 0$.  That is $x\ge -2$.  Then

$6 -(x+2) = px + 5$
$4 -x = px + 5$
$-1 = px + x= x(p+1)$
$x = -\frac 1{p+1}$.  But this can only be a solution if a) $p+1\ne 0$ that is $p \ne -1$ and b) $x \ge -2$.  And if $x =-\frac 1{p+1} \ge -2$ then
$\frac 1{p+1} \le 2$
If $p< -1$ then $\frac 1{p+1} < 0 < 2$ so this will always be true.
But if $p > -1$ then we need $0< \frac 1{p+1} < 2$ or $1 < 2(p+1)$ so $p>-\frac 12$.

$x =-\frac 1{p+1}$ will be one solution if $p < -1$ or $p>-\frac 12$.



*$x+2 < 0$. That is $x < -2$. then

$6 -(-(x+2)) = px + 5$
$6 + x+2 = px+5$
$8 + x= px + 5$
$3 = px -x= x(p-1)$
$x = \frac 3{p-1}$
But this can only be a solution if a) $p-1\ne 0$ or $p =1$.  and $b) x = \frac 3{p+1} < -2$.
If $p > 1$ then $\frac 3{p-1} > 0 > -2$ and this can never happen.
If $p < 1$ then we have $\frac 3{p-1} < 0$ and we need
$\frac 3{p-1} < -2$
$3 > -2(p-1)= -2p +2$
$1 > -2p$
$p > -\frac 12$.

$x = \frac 3{p-1}$ will be one solution if $-\frac 12 < p < 1$.

Putting those two results together:

if $p < -1$  then $x=-\frac 1{p+1}$ will be the only solution.
if $-1 \le p \le -\frac12$ then there will be no solutions.
if $-\frac 12 < p < 1$ there there will be two solutions.  $x =-\frac 1{p+1}$ and $x = \frac 3{p-1}$.
if $x \ge 1$ then $x=-\frac 1{p+1}$ will be the only solution.

A: Let $$g(x):=6-|x+2|-(px+5)=1-|x+2|-px.$$
You are asking the roots of this function.
Notice that it is piecewise linear, i.e. monotonous between the "corner" points, and it suffices to detect the changes of signs between these corner points and the points at $\pm\infty$.
We see that

*

*$g(-\infty)$ has the sign of $p-1$,


*$g(-2)=1+2p$,


*$g(\infty)$ has the sign of $-p-1$.
Now the plot below shows you the values of these functions (in the order blue, green, magenta) and you just have to count the changes of sign.


 Blue and green differ in sign between $-\frac12$ and $1$, while green and magenta differ everywhere but between $-1$ and $-\frac12$.

A: The solution given by @fleablood is exact.
In such a case, I think an instructor can as well accept a graphical solution because

*

*the algebraic separation of cases is very tedious.


*the ability to "extract information" from a graphics is often a testimony of good understanding.
If such is the case, everything can be deduced from the observation of the following graphics with the "critical" slopes $m = -1,-1/2,1.$

