Question

y = 2·4^x
show your work here
draw
14
12
10
8
6
4
2

Explanation:

Step1: Choose x-values

Pick \( x = 0, 1, 2 \) (simple values for calculation).

Step2: Calculate y for \( x = 0 \)

Substitute \( x = 0 \) into \( y = 2 \cdot 4^x \):
\( y = 2 \cdot 4^0 = 2 \cdot 1 = 2 \). So point: \( (0, 2) \).

Step3: Calculate y for \( x = 1 \)

Substitute \( x = 1 \):
\( y = 2 \cdot 4^1 = 2 \cdot 4 = 8 \). So point: \( (1, 8) \).

Step4: Calculate y for \( x = 2 \)

Substitute \( x = 2 \):
\( y = 2 \cdot 4^2 = 2 \cdot 16 = 32 \) (but 32 > 14, so maybe \( x = -1 \) instead).
For \( x = -1 \): \( y = 2 \cdot 4^{-1} = 2 \cdot \frac{1}{4} = 0.5 \). Point: \( (-1, 0.5) \).

Step5: Plot points

Plot \( (-1, 0.5) \), \( (0, 2) \), \( (1, 8) \). The graph is an exponential curve, increasing (since base \( 4 > 1 \)) with y-intercept at \( (0, 2) \), passing through \( (1, 8) \), and approaching the x-axis as \( x \to -\infty \).

Answer:

To graph \( y = 2 \cdot 4^x \), follow these steps:

1. Identify the function type: Exponential growth (base \( 4 > 1 \), coefficient \( 2 > 0 \)).
2. Find key points:

• At \( x = 0 \): \( y = 2 \cdot 4^0 = 2 \) (point: \( (0, 2) \), y-intercept).
• At \( x = 1 \): \( y = 2 \cdot 4^1 = 8 \) (point: \( (1, 8) \)).
• At \( x = -1 \): \( y = 2 \cdot 4^{-1} = 0.5 \) (point: \( (-1, 0.5) \)).

1. Plot points and draw a smooth curve through them, approaching the x-axis as \( x \to -\infty \) and increasing rapidly as \( x \to \infty \).

(Note: The pre-drawn blue line in the image likely represents the horizontal asymptote \( y = 0 \), consistent with exponential functions \( a \cdot b^x \) ( \( b > 0, b
eq 1 \)) having \( y = 0 \) as an asymptote.)