Question

graph $f(x) = 2 \cdot 3^{x + 1} + 1$

$x$	$f(x)$
-1
0
1
2

domain:

range:

y-intercept:

equation of asymptote:

Explanation:

Step1: Fill the table with x values

For \( x = -2 \): \( f(-2)=2\cdot3^{-2 + 1}+1 = 2\cdot3^{-1}+1=\frac{2}{3}+1=\frac{5}{3}\approx1.67 \)
For \( x = -1 \): \( f(-1)=2\cdot3^{-1 + 1}+1 = 2\cdot3^{0}+1=2 + 1 = 3 \)
For \( x = 0 \): \( f(0)=2\cdot3^{0 + 1}+1 = 2\cdot3^{1}+1=6 + 1 = 7 \)
For \( x = 1 \): \( f(1)=2\cdot3^{1 + 1}+1 = 2\cdot3^{2}+1=18 + 1 = 19 \)
For \( x = 2 \): \( f(2)=2\cdot3^{2 + 1}+1 = 2\cdot3^{3}+1=54 + 1 = 55 \)

Step2: Determine Domain

The function \( f(x)=2\cdot3^{x + 1}+1 \) is an exponential function. Exponential functions have domain all real numbers. So Domain: \( (-\infty,\infty) \) or all real numbers.

Step3: Determine Range

For exponential function \( a\cdot b^{x + c}+d \), when \( b>1 \), as \( x\to-\infty \), \( b^{x + c}\to0 \). Here \( a = 2 \), \( b = 3 \), \( d = 1 \). So \( \lim_{x\to-\infty}f(x)=2\cdot0 + 1 = 1 \). As \( x\to\infty \), \( 3^{x+1}\to\infty \), so \( f(x)\to\infty \). So Range: \( (1,\infty) \)

Step4: Determine y - intercept

Y - intercept is when \( x = 0 \). From step 1, when \( x = 0 \), \( f(0)=7 \). So y - intercept is \( 7 \) (or the point \( (0,7) \))

Step5: Determine Asymptote

For exponential function \( f(x)=a\cdot b^{x + c}+d \), the horizontal asymptote is \( y = d \). Here \( d = 1 \), so the equation of the asymptote is \( y = 1 \)

Answer:

Filled Table:

\( x \)	\( f(x) \)
-1	3
0	7
1	19
2	55

Domain: \( (-\infty,\infty) \) (All real numbers)

Range: \( (1,\infty) \)

y - intercept: \( 7 \) (or \( (0,7) \))

Equation of Asymptote: \( y = 1 \)