1. The problem asks to find two points with integer coordinates on the graph of the function $f(x) = 3 \left(\frac{1}{2}\right)^x - 5$. 2. The function is an exponential decay function shifted down by 5 units. 3. To find points, substitute integer values of $x$ and calculate $f(x)$. 4. For $x=0$: $$f(0) = 3 \left(\frac{1}{2}\right)^0 - 5 = 3 \cdot 1 - 5 = 3 - 5 = -2$$ So the point is $(0, -2)$. 5. For $x=1$: $$f(1) = 3 \left(\frac{1}{2}\right)^1 - 5 = 3 \cdot \frac{1}{2} - 5 = \frac{3}{2} - 5 = \frac{3}{2} - \frac{10}{2} = -\frac{7}{2} = -3.5$$ This is not an integer coordinate for $y$. 6. For $x=2$: $$f(2) = 3 \left(\frac{1}{2}\right)^2 - 5 = 3 \cdot \frac{1}{4} - 5 = \frac{3}{4} - 5 = -\frac{17}{4} = -4.25$$ Not integer. 7. For $x=-1$: $$f(-1) = 3 \left(\frac{1}{2}\right)^{-1} - 5 = 3 \cdot 2 - 5 = 6 - 5 = 1$$ So the point is $(-1, 1)$. 8. The two points with integer coordinates are $(-1, 1)$ and $(0, -2)$. 9. The horizontal asymptote is $y = -5$. Final answer: The two points are $(-1, 1)$ and $(0, -2)$.