Finding $a$ such that $f(x)=a\,|x-b|$ is satisfied by $(2,1)$ and $(10,3)$ 
I was given the following equation: $$f(x)=a\,|x-b|$$ and the information that both $(2,1)$ and $(10,3)$ are solutions for the equation. The question asked to solve for $a$.

This is what I did: $$a=\frac{1}{2-b}\\[10pt]a=\frac{-1}{2-b}\\[10pt]a=\frac{-3}{10-b}\\[10pt]a=\frac{3}{10-b}$$ all of these seem to be possible solutions for the equation, I just need to find the one that will work with both solution sets. I solved and got multiple values for a and b. The ones that worked for both ordered pairs were the solutions $a=\frac12 \text{ and } b=4$ and $a=\frac14\text{ and } b=-2$.
I feel pretty confident about my answer (perhaps wrongly so :) ) but is there a faster, more efficient way to do this?
 A: Your answers are correct, but there is a more efficient way of finding them using what we know about the vertex in this context.
First, note that $f(x) = a|x-b|$ has two transformations, a shift to the right by $b$ and a vertical stretch by $a$. So, we know that the vertex has coordinates $(b,0)$. Since the vertex must be on the $x$-axis, there are only two possible cases here, each with exactly one possible solution: either the vertex is left of $x = 2$ or it's between $x = 2$ and $x = 10$.
Case 1: $b<2$
If the vertex is left of $x = 2$, we know that $(b,0)$, $(2,1)$, and $(10,3)$ are collinear with slope $a$. We start by finding the slope between $(2,1)$ and $(10,3)$:
$$ a = \frac{3-1}{10-2} = \frac{1}{4} . $$
Then, we use $a$ to find $b$:
$$ \frac{1-0}{2-b} = \frac{1}{4} \Rightarrow 2-b = 4 \Rightarrow b = -2 . $$
The first solution is $a = \frac{1}{4}, b = -2$.
Case 2: $ 2 < b < 10$
Here, we use the absolute value's vertical symmetry about the vertex - the slopes must be opposites on either side of the vertex:
$$ \frac{0-1}{b-2} = -\frac{3-0}{10-b} \Rightarrow b-10=3(b-2) \Rightarrow b = 4 . $$
Then, $a$ is the slope to the right of the vertex:
$$ a = \frac{3-0}{10-b} = \frac{3}{10-4} = \frac{1}{2} . $$
The second solution is $a = \frac{1}{2}, b = 4$.
A: If $a=0$ we have the constant function $f(x) = 0$ and we are done.
This is not our case, as we have solution points given which are not on the $x$-axis.
Thus $a\ne 0$. The given function $f(x) = a \lvert x-b\rvert$ is a variation of the absolute value, thus V-shaped, if $a$ is positive, the tip somewhere on the $x$-axis.
For the two given points $P=(2,1)$ and $Q=(10,3)$, we have two cases:

*

*$P$ and $Q$ lie on the same arm of the V

*$P$ and $Q$ lie on two different arms of the V

Case 1:
$f$ is growing from $P$ to $Q$, so the tip $(b, 0)$ must lie to the left of $P$ and $Q$. We can drop the absolute value and get the system of equations:
\begin{align}
1 = f(2) &= a (2 - b) = 2 a - ab \\
3 = f(10) &= a (10 - b) = 10 a - ab
\end{align}
Subtraction of the equations gives
$$
2 = 8 a \iff a = 1/4
$$
Inserting this in the first equation gives:
$$
1 = 1/2 - b/4 \iff 4 = 2 - b \iff b = -2
$$
Case 2:
We have the tip in between $P$ and $Q$, which means the absolute value is resolved for the negative case for the first equation and we get the system:
\begin{align}
1 = f(2) &= a (b - 2) = -2 a + ab \\
3 = f(10) &= a (10 - b) = 10 a - ab
\end{align}
Addition gives
$$
4 = 8 a \iff a = 1/2
$$
Inserting this in the first equation gives
$$
1 = -1 + b/2 \iff 2 = -2 + b \iff b = 4
$$
A: I see two cases.  $b\in(2,10)$ and $b\notin(2,10)$
$f(b) = 0$
In the case that $b$ is not in $(2,10)$ we construct the line through the two points.    $b$ is the $x$-intercept of this line.  $|a|$ is the slope.
$|a| = \frac {3-1}{10-2} = \frac 14$
Since this slope is postive, $a > 0, b < 2$
$0 = \frac 14 (x-2) + 1\\
0 = \frac 14 (x +2)\\
b = -2\\
f(x) = \frac 14 |x+2|$
In the case that $b\in(2,10)$  The line needs to drop one unit in the interval and then rise 3 units.
$a = \frac {3+1}{10-2} = \frac 12$
And we have a negative slope to the left of $b$ and a postive slope to the right.
$0 = \frac 12 (x-10) + 3\\
0 = \frac 12 (x-4)
b = 4$
or
$0 = \frac 12 (x-2) + 1\\
0 = \frac 12 (x-4)
b = 4$
$f(x) = \frac 12 |x-4|$
A: $${1\over a}=|2-b|\tag1$$
$${3\over a}=|10-b|\tag2$$
By (1) and (2),
$${{10-b}\over{2-b}}=\pm3\tag3$$
Solving (3), $b=-2$ (for $+3$ on right) or $b=4.$
By (1), $a$ is $1\over4$ or $1\over2$.
