Question

1. graph f(x) = 3(3^x) + 1

describe the transformations that made f(x).

write the following values for f(x):
y-intercept:
horizontal asymptote:
domain:
range:

Explanation:

Step1: Recall Parent Function

The parent function is \( y = 3^x \), which has a horizontal asymptote at \( y = 0 \), domain \( (-\infty, \infty) \), range \( (0, \infty) \), and y - intercept at \( x = 0,y = 1 \).

Step2: Analyze Transformations

• For the function \( f(x)=3(3^{x}) + 1=3^{x + 1}+1 \) (using the property \( a\times a^{x}=a^{x + 1} \)):
• The coefficient \( 3\) in front of \( 3^{x} \) is a vertical stretch by a factor of \( 3 \) (since \( a>1\) in \( y = a\times b^{x} \) for vertical stretch). Also, we can rewrite \( 3\times3^{x}=3^{x + 1} \), which is a horizontal shift? Wait, no. Wait, \( y = 3\times3^{x}=3^{x+1} \) is a horizontal shift? No, \( a\times b^{x}=b^{x+\log_{b}a} \). For \( b = 3,a = 3 \), \( \log_{3}3=1 \), so \( 3\times3^{x}=3^{x + 1} \), which is a horizontal shift left by \( 1 \) unit? Wait, no, the standard transformation: if we have \( y = b^{x + h} \), it's a horizontal shift left by \( h \) units. But also, \( y = a\times b^{x} \) is a vertical stretch by factor \( a \) when \( a>0 \). Let's do it step by step.
• Start with \( y = 3^{x} \). Multiply by \( 3 \): \( y = 3\times3^{x} \), this is a vertical stretch by a factor of \( 3 \) (since we multiply the function value by \( 3 \) for each \( x \)). Then add \( 1 \): \( y=3\times3^{x}+1 \), this is a vertical shift up by \( 1 \) unit.

Step3: Find y - intercept

To find the y - intercept, set \( x = 0 \):
\( f(0)=3(3^{0})+1=3\times1 + 1=3 + 1 = 4 \)

Step4: Find Horizontal Asymptote

For exponential functions of the form \( y = a\times b^{x}+k \), the horizontal asymptote is \( y = k \). Here, the function is \( f(x)=3\times3^{x}+1 \), as \( x
ightarrow-\infty \), \( 3^{x}
ightarrow0 \), so \( 3\times3^{x}
ightarrow0 \), and \( f(x)
ightarrow0 + 1=1 \). So the horizontal asymptote is \( y = 1 \).

Step5: Find Domain

Exponential functions of the form \( y = a\times b^{x}+k \) (where \( a
eq0,b>0,b
eq1 \)) have domain \( (-\infty,\infty) \) because we can plug in any real number for \( x \). So the domain of \( f(x)=3\times3^{x}+1 \) is \( (-\infty,\infty) \).

Step6: Find Range

As \( x
ightarrow-\infty \), \( 3^{x}
ightarrow0 \), so \( 3\times3^{x}
ightarrow0 \), and \( f(x)
ightarrow0 + 1 = 1 \). As \( x
ightarrow\infty \), \( 3^{x}
ightarrow\infty \), so \( 3\times3^{x}
ightarrow\infty \), and \( f(x)
ightarrow\infty+1=\infty \). Since the function is increasing (because the base \( 3>1 \) and we have a vertical stretch and shift up, which doesn't change the increasing nature), the range is \( (1,\infty) \).

Answer:

Transformations:

Vertical stretch by a factor of \( 3 \) of the parent function \( y = 3^{x} \), followed by a vertical shift up by \( 1 \) unit. (Or we can think of it as a horizontal shift left by \( 1 \) unit (since \( 3\times3^{x}=3^{x + 1} \)) followed by a vertical shift up by \( 1 \) unit)

y - intercept:

\( 4 \)

Horizontal Asymptote:

\( y = 1 \)

Domain:

\( (-\infty, \infty) \)

Range:

\( (1, \infty) \)