Question

the graph of $y = 3^x$ is shown below.
graph of $y = 3^x$
which of the following is the graph of $y = 3^{x + 3} + 3$?
choose 1 answer:
a graph a
b graph b
c graph c
d graph d

Explanation:

Step1: Analyze horizontal shift

For the function \( y = 3^{x + 3}+3 \), compared to \( y = 3^{x} \), the \( x+3 \) inside the exponent means a horizontal shift. The rule for horizontal shifts is: if we have \( y = f(x + h) \), it's a shift of \( h \) units to the left (when \( h>0 \)). Here, \( h = 3 \), so the graph of \( y = 3^{x} \) shifts 3 units to the left.

Step2: Analyze vertical shift

The \( +3 \) at the end means a vertical shift. The rule for vertical shifts is: if we have \( y = f(x)+k \), it's a shift of \( k \) units up (when \( k>0 \)). Here, \( k = 3 \), so the graph shifts 3 units up.

Step3: Check the y - intercept (optional but helpful)

For \( y = 3^{x} \), when \( x = 0 \), \( y=3^{0}=1 \). For \( y = 3^{x + 3}+3 \), when \( x=- 3 \), \( y = 3^{-3 + 3}+3=3^{0}+3=1 + 3=4 \)? Wait, no, let's find the y - intercept (x = 0) for \( y=3^{x + 3}+3 \). When \( x = 0 \), \( y=3^{0 + 3}+3=27 + 3=30 \)? Wait, maybe better to look at the horizontal asymptote. The horizontal asymptote of \( y = 3^{x} \) is \( y = 0 \). For \( y=3^{x+3}+3 \), the horizontal asymptote is \( y=3 \) (since the \( +3 \) shifts the asymptote up 3 units) and the graph is shifted 3 units left.

Looking at the options:
- Option A: Horizontal asymptote seems to be around \( y = 3 \) (vertical shift up 3) and shifted left? Wait, no, let's re - evaluate. Wait, the original function \( y = 3^{x} \) has a y - intercept at (0,1). After shifting left 3 units and up 3 units: the point (0,1) on \( y = 3^{x} \) moves to \( (0 - 3,1 + 3)=(-3,4) \). Now let's check the options:
- Option B: The graph seems to have a horizontal asymptote and a shift that doesn't match.
- Option C: The horizontal asymptote is \( y = 3 \) (vertical shift up 3) and the graph is shifted left (since the "start" of the exponential growth is shifted left) and the shape. Wait, no, let's think again. Wait, the function \( y=3^{x+3}+3=3^{3}\times3^{x}+3 = 27\times3^{x}+3 \). But the key shifts: left 3, up 3. The horizontal asymptote is \( y = 3 \), and the graph is an exponential growth shifted left 3 and up 3. Looking at the options, the graph with horizontal asymptote \( y = 3 \) (so the graph approaches \( y = 3 \) as \( x ightarrow-\infty \)) and shifted left (so the "rise" of the exponential happens earlier (more to the left)) and up 3. Option C has a horizontal asymptote around \( y = 3 \) and the shape consistent with left shift and up shift. Wait, maybe I made a mistake earlier. Wait, the correct analysis:

The parent function \( y = 3^{x} \): domain \( \mathbb{R} \), range \( (0,\infty) \), horizontal asymptote \( y = 0 \), increasing, passes through (0,1).

For \( y=3^{x + 3}+3 \):

- Horizontal asymptote: As \( x ightarrow-\infty \), \( 3^{x+3} ightarrow0 \), so \( y ightarrow3 \). So horizontal asymptote \( y = 3 \).
- When \( x=-3 \), \( y=3^{-3 + 3}+3=1 + 3=4 \).
- The graph is obtained by shifting \( y = 3^{x} \) 3 units left (so the point (0,1) on \( y = 3^{x} \) goes to (- 3,1) and then up 3 units to (-3,4)) and 3 units up.

Now looking at the options:

- Option A: The horizontal asymptote is around \( y = 3 \), but the shape of the exponential part: when we shift left 3 and up 3, the exponential growth should start (in terms of the "steep" part) more to the left. Wait, maybe the correct answer is C? Wait, no, let's check the options again. Wait, maybe I messed up the shift direction. Wait, \( y = f(x + h) \) is left shift, \( y = f(x - h) \) is right shift. So \( y=3^{x+3}=3^{x-(-3)} \), so it's a shift 3 units to the left. And \( +3 \) is up 3 units.

Looking at the options, the graph with horizontal asymptote \( y = 3 \) (so the lowest the graph goes is approaching 3 from above) and shifted left (so the exponential curve starts to rise more to the left) and up 3. The correct option should be the one where the horizontal asymptote is \( y = 3 \) and the graph is shifted left and up. From the given options, option C has a horizontal asymptote at \( y = 3 \) (the graph approaches \( y = 3 \) as \( x ightarrow-\infty \)) and the shape of the exponential function shifted left and up. Wait, maybe the answer is C? Wait, no, let's check the original graph of \( y = 3^{x} \) (given) has a y - intercept at (0,1) (since when x = 0, y = 1). For \( y=3^{x+3}+3 \), when x=-3, y=4 (as 3^0 + 3=4), and the horizontal asymptote is y = 3. Looking at the options, option C has a horizontal asymptote at y = 3 and the graph is shifted left (so the exponential growth is shifted left) and up. So the correct answer is C? Wait, no, maybe I made a mistake. Wait, let's re - check the options:

Wait, the original function \( y = 3^{x} \) is an exponential growth with y - intercept (0,1). The function \( y=3^{x + 3}+3 \) is \( y = 3^{3}\times3^{x}+3=27\times3^{x}+3 \). So it's a steeper exponential (since 27>1) shifted up 3 units? Wait, no, \( 3^{x+3}=3^{3}\times3^{x} \), so it's a vertical stretch by a factor of 27 and a vertical shift up 3? Wait, no, the horizontal shift is 3 units left, and vertical shift up 3. The vertical stretch is a result of the \( 3^{x+3}=3^{3}\times3^{x} \), but the main shifts are left 3 and up 3.

Looking at the options:

- Option A: The horizontal asymptote is y = 3, but the exponential part seems to have a y - intercept (at x = 0) that is too low.
- Option B: The horizontal asymptote is not y = 3.
- Option C: The horizontal asymptote is y = 3, and the graph is shifted left (so the exponential growth starts more to the left) and up 3.
- Option D: The horizontal asymptote is not y = 3.

So the correct answer should be C? Wait, no, maybe I made a mistake in the horizontal shift. Wait, \( y = 3^{x+3} \) is a shift of \( y = 3^{x} \) 3 units to the left, and then \( +3 \) is 3 units up. So the graph of \( y = 3^{x} \) (which has a horizontal asymptote at y = 0, passes through (0,1)) when shifted left 3 units, the point (0,1) goes to (-3,1), then shifted up 3 units to (-3,4). The horizontal asymptote goes from y = 0 to y = 3. Now looking at the options, option C has a horizontal asymptote at y = 3 and the graph is an exponential growth shifted left (so the "curve" starts to rise when x is more negative) and up 3. So the answer is C.

Answer:

C