Question

describe how the graph of $y = 2 \cdot 3^x - 1$ compares to its parent function. (1 point) \
\bigcirc it is the graph of $y = 3^x$ with a vertical stretch by a factor of 2 and shifted right 1 unit. \
\bigcirc it is the graph of $y = 3^x$ with a vertical stretch by a factor of 2 and shifted down 1 unit. \
\bigcirc it is the graph of $y = 3^x$ with a horizontal stretch by a factor of 2 and shifted down 1 unit. \
\bigcirc it is the graph of $y = 3^x$ with a vertical compression by a factor of $\frac{1}{2}$ and shifted down 1 unit.

Explanation:

1. Recall the transformations of exponential functions. For a function \( y = a\cdot b^{x}+k \), where \( y = b^{x} \) is the parent function:

• The coefficient \( a \) affects vertical stretch/compression: if \( |a|>1 \), it's a vertical stretch by a factor of \( |a| \); if \( 0<|a|<1 \), it's a vertical compression.
• The constant \( k \) affects vertical shift: if \( k>0 \), shift up \( k \) units; if \( k < 0 \), shift down \( |k| \) units.

1. In the function \( y = 2\cdot3^{x}-1 \), compared to the parent function \( y = 3^{x} \):

• The coefficient \( a = 2 \) (since \( |2|>1 \)), so there is a vertical stretch by a factor of 2.
• The constant \( k=-1 \), so the graph is shifted down 1 unit (because we subtract 1 from the function \( 2\cdot3^{x} \)).
• There is no horizontal stretch here (horizontal stretch would involve a change in the exponent's coefficient, like \( y = 3^{bx} \), which is not the case here). Also, a vertical compression would be when \( |a|<1 \), but \( a = 2>1 \), so it's a stretch, not a compression.
• Shifting right would involve a change like \( y = 3^{x - h} \) with \( h>0 \), which is not present here.

Answer:

B. It is the graph of \( y = 3^{x} \) with a vertical stretch by a factor of 2 and shifted down 1 unit.