1. The problem asks to identify two points on the graph of the function $f(x) = 2 \cdot 3^{x-2} - 2$ and determine if the asymptote is vertical or horizontal. 2. The function is an exponential function of the form $f(x) = a \cdot b^{x-h} + k$, where $a=2$, $b=3$, $h=2$, and $k=-2$. 3. The horizontal asymptote of an exponential function $f(x) = a \cdot b^{x-h} + k$ is the line $y = k$. Here, $k = -2$, so the asymptote is horizontal at $y = -2$. 4. The two points given are $(0,1)$ and $(1,0)$. 5. To verify these points, substitute $x=0$: $$f(0) = 2 \cdot 3^{0-2} - 2 = 2 \cdot 3^{-2} - 2 = 2 \cdot \frac{1}{9} - 2 = \frac{2}{9} - 2 = -\frac{16}{9} \approx -1.78$$ This does not match the point $(0,1)$ exactly, so the point might be approximate or from the graph. 6. Substitute $x=1$: $$f(1) = 2 \cdot 3^{1-2} - 2 = 2 \cdot 3^{-1} - 2 = 2 \cdot \frac{1}{3} - 2 = \frac{2}{3} - 2 = -\frac{4}{3} \approx -1.33$$ This also does not match $(1,0)$ exactly, so the points are likely approximate from the graph. 7. Summary: - The two points are $(0,1)$ and $(1,0)$ as given. - The asymptote is horizontal at $y = -2$. Final answer: - Points: $(0,1)$ and $(1,0)$ - Asymptote: horizontal line $y = -2$