my question is how can i get a equation out of this i have gathered the following
$$ y = a|x-h| + k$$
$$y = a|x-1| - 4$$ ($a$ i don't know yet and this is where i have a question)
to find $a$ can i just use any point in the graph and plug it in the equation or is their any other ways of doing it?
On
This is $$y = a|x-1| - 4$$ your equation. It's a translation of the modulus function by 1 unit in the positive $x$ direction and by 4 units in the negative $y$ direction. The gradient is given by $a$.
You can see it's in the form $y=mx+c$ with $m=a$, $c=-4$ and $x=\mid x-1 \mid$ but this is the equation for a straight line. However, the modulus takes two values, namely, $x-1$ and $1-x$, so there will be 2 straight line equations.
When dealing with modulus functions you must consider the two separate cases to find the component lines.
So the equation of the left line is $\color{blue}{y=a(1-x) -4}$ and the right line has equation $\color{red}{y=a(x-1) -4}$.
$a$ can be found by substituting in a point that lies on the left line; for example substitute $(-1,0)$ into the $\color{blue}{\mathrm{blue}}$ equation and solve for $a$:
$$\color{blue}{0=a(1--1) -4}$$ $$\implies \color{blue}{0=2a -4}$$ $$\implies \color{blue}{a=2}$$
Substituting $(3,0)$ into the $\color{red}{\mathrm{red}}$ equation gives: $$\color{red}{0=a(3-1) -4}$$ $$\implies \color{red}{0=2a -4}$$ $$\implies \color{red}{a=2}$$
as before.
It seems that in your equation $a=2$ to make the equation $y=2|x−1|−4$.
Verify this from the points: $(1, -4)$, $(3,0)$, and $(-1, 0)$.
$a$ is the slope of your equation.