The zeroes of the function $y = f(x)$ are $a, b, c.$ What are the new zeroes if that function is transformed into $y = \dfrac{1}{2f(-rx - t)}?$
My solution: $(a, 0) , (b, 0) , (c, 0)$.
$1)$ Reflection about the $y$ - axis: $(-a, 0) , (-b, 0) , (-c, 0)$
$2)$ Vertical compression by a factor of $1/2$: no affect $(-a, 0) , (-b, 0) , (-c, 0)$
$3)$ Horizontal compression by a factor of 1/r: $(-a/r, 0) , (-b/r, 0) , (-c/r, 0)$
$4)$ Horizontal translation t/r units left: $(-a/r - t/r, 0) , (-b/r - t/r, 0) , (-c/r - t/r, 0)$
Am I correct?
zeroes of the newly transformed function are : $$ ((-a/r - t/r), 0), \space (-(b/r - t/r), 0), \space (-(c/r - t/r), 0) $$
You are correct.
Generally, suppose $g(x)=f(rx+t).$ Then $g(x)=0$ iff $f(rx+t)=0$ iff $rx+t=z$ for $z$ any zero of $f$. Then the zeros of $g$ are given by $\frac{z-t}{r}$ for $z$ any zero of $f$.