The problem states that the function $f(x) = log_{10}x$ has the point $(10, 1)$ on its graph. We need to find the coordinates of the transformed point from $(10, 1)$ on the transformed function $f(x) = 3log_{10}(-\frac{1}{2}(x+4))-6$.

1. Problem Description

The problem states that the function $f(x) = log_{10}x$ has the point $(10, 1)$ on its graph. We need to find the coordinates of the transformed point from $(10, 1)$ on the transformed function $f(x) = 3log_{10}(-\frac{1}{2}(x+4))-6$.

2. Solution Steps

Let the original point be $(x, y) = (10, 1)$.

The transformed function is given by $f(x) = 3log_{10}(-\frac{1}{2}(x+4)) - 6$.

The general transformation is given by $y' = af(b(x'-c))+d$.

Comparing with the transformed function $f(x) = 3log_{10}(-\frac{1}{2}(x+4))-6$, we have:

$a = 3$, $b = -\frac{1}{2}$, $c = -4$, $d = -6$.

We have the transformation equations:

$y = af(b(x' - c)) + d$, where $x=x'$ and $y=f(x)$.

For the x-coordinate:

$x = b(x' - c)$ implies that $x' = \frac{x}{b} + c$.

In this case, we have $x = 10$, $b = -\frac{1}{2}$, and $c = -4$.

Therefore, $x' = \frac{10}{-\frac{1}{2}} + (-4) = -20 - 4 = -24$.

For the y-coordinate:

$y' = ay + d$.

In this case, we have $y = 1$, $a = 3$, and $d = -6$.

Therefore, $y' = 3(1) + (-6) = 3 - 6 = -3$.

So the transformed point is $(-24, -3)$.

3. Final Answer

(-24, -3)