The graph of $$ g(x)=a\left(\frac{1}{5}\right)^{x}-5 $$ passes throunti $$ (-2 ;-4) $$. 7.1 Show that $$ \alpha=\frac{1}{25} $$ 7.2 Determine the coordinates of the $$ x $$ intericentof 7.3 Given: $$ h(x)=\left(\frac{1}{5}\right)^{x} $$ 7.3.1 Determine the equation of $$ x^{-1}(x) $$, the inverse of $$ h $$, in the for $$ y=\ldots $$ 7.3.2 Describe the transformation from. g to h .

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### 7.1 Show that $$ \alpha = \frac{1}{25} $$

We know that the graph of $$ g(x) = a\left(\frac{1}{5}\right)^{x} - 5 $$ passes through the point $$ (-2, -4) $$. This means that when $$ x = -2 $$, $$ g(-2) = -4 $$.

Substituting $$ x = -2 $$ into the equation:

$$
g(-2) = a\left(\frac{1}{5}\right)^{-2} - 5 = -4
$$

Calculating $$ \left(\frac{1}{5}\right)^{-2} $$:

$$
\left(\frac{1}{5}\right)^{-2} = 5^2 = 25
$$

Now substituting this back into the equation:

$$
g(-2) = a \cdot 25 - 5 = -4
$$

Rearranging gives:

$$
a \cdot 25 = -4 + 5
$$
$$
a \cdot 25 = 1
$$
$$
a = \frac{1}{25}
$$

Thus, we have shown that $$ \alpha = \frac{1}{25} $$.

### 7.2 Determine the coordinates of the $$ x $$-intercept

The $$ x $$-intercept occurs when $$ g(x) = 0 $$. We can set the equation to zero and solve for $$ x $$:

$$
0 = \frac{1}{25}\left(\frac{1}{5}\right)^{x} - 5
$$

Rearranging gives:

$$
\frac{1}{25}\left(\frac{1}{5}\right)^{x} = 5
$$

Multiplying both sides by 25:

$$
\left(\frac{1}{5}\right)^{x} = 125
$$

Since $$ 125 = 5^3 $$, we can rewrite this as:

$$
\left(\frac{1}{5}\right)^{x} = 5^{-3}
$$

This implies:

$$
-x = -3 \quad \Rightarrow \quad x = 3
$$

Thus, the coordinates of the $$ x $$-intercept are $$ (3, 0) $$.

### 7.3 Given: $$ h(x) = \left(\frac{1}{5}\right)^{x} $$

#### 7.3.1 Determine the equation of $$ x^{-1}(x) $$, the inverse of $$ h $$

To find the inverse of $$ h(x) $$, we start with:

$$
y = \left(\frac{1}{5}\right)^{x}
$$

Now, we solve for $$ x $$:

1. Take the logarithm of both sides:
   $$
   \log(y) = x \log\left(\frac{1}{5}\right)
   $$

2. Rearranging gives:
   $$
   x = \frac{\log(y)}{\log\left(\frac{1}{5}\right)}
   $$

Thus, the inverse function is:

$$
x^{-1}(y) = \frac{\log(y)}{\log\left(\frac{1}{5}\right)}
$$

In terms of $$ y $$:

$$
y = \left(\frac{1}{5}\right)^{x} \quad \Rightarrow \quad x = -\log_{5}(y)
$$

So, the equation of the inverse is:

$$
y = -\log_{5}(x)
$$

#### 7.3.2 Describe the transformation from $$ g $$ to $$ h $$

The transformation from $$ g(x) $$ to $$ h(x) $$ can be described as follows:

1. **Vertical Stretch/Compression**: The function $$ g(x) $$ is vertically compressed by a factor of $$ \frac{1}{25} $$ compared to $$ h(x) $$.
2. **Vertical Shift**: The function $$ g(x) $$ is shifted down by 5 units compared to $$ h(x) $$.

In summary, the transformation from $$ h(x) $$ to $$ g(x) $$ involves a vertical compression by a factor of $$ \frac{1}{25} $$ and a downward shift of 5 units.
7.1 $$ \alpha = \frac{1}{25} $$ 7.2 The $$ x $$-intercept is at $$ (3, 0) $$. 7.3.1 The inverse of $$ h(x) $$ is $$ y = -\log_{5}(x) $$. 7.3.2 The transformation from $$ g(x) $$ to $$ h(x) $$ involves a vertical compression by a factor of $$ \frac{1}{25} $$ and a downward shift of 5 units.

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