Exponential Transformation 8Be5B9

1. **State the problem:** We need to graph the function $$y = \frac{1}{8} \cdot 2^x$$ by transforming the parent function $$y = 2^x$$. 2. **Recall the parent function:** The parent function is $$y = 2^x$$, which is an exponential growth function with a horizontal asymptote at $$y=0$$. 3. **Transformation rule:** Multiplying the parent function by $$\frac{1}{8}$$ vertically compresses the graph by a factor of $$\frac{1}{8}$$. 4. **Apply the transformation:** - The original points on $$y=2^x$$ include: - At $$x=0$$, $$y=2^0=1$$ - At $$x=3$$, $$y=2^3=8$$ - At $$x=6$$, $$y=2^6=64$$ - At $$x=9$$, $$y=2^9=512$$ - After multiplying by $$\frac{1}{8}$$, the points become: - At $$x=0$$, $$y=\frac{1}{8} \cdot 1 = \frac{1}{8}$$ - At $$x=3$$, $$y=\frac{1}{8} \cdot 8 = 1$$ - At $$x=6$$, $$y=\frac{1}{8} \cdot 64 = 8$$ - At $$x=9$$, $$y=\frac{1}{8} \cdot 512 = 64$$ 5. **Graph shape:** The graph is still an increasing exponential growth curve with the same horizontal asymptote $$y=0$$, but it is vertically compressed. 6. **Summary:** The transformation $$y = \frac{1}{8} \cdot 2^x$$ compresses the parent graph vertically by a factor of $$\frac{1}{8}$$, shifting points accordingly. **Final answer:** The graph passes through points approximately $$(0, \frac{1}{8})$$, $$(3, 1)$$, $$(6, 8)$$, and $$(9, 64)$$ with horizontal asymptote $$y=0$$.