Question

without graphing, complete the following for the function k(x)=3(2^x).
(a) describe the shape of the graph of the function.
(b) find the second coordinates of the points with first coordinates 0 and 1.
(a) the graph is entirely
dropdown the x - axis and
dropdown from left to right,
dropdown steeply than the graph of y = 2^x.
options for the first dropdown: above, below
options for the second dropdown: (presumably related to increasing/decreasing, but text not fully clear)
options for the third dropdown: (presumably related to how steeply)

Explanation:

Step1: Analyze the exponential function shape

The function $k(x)=3(2^x)$ is a positive vertical scaling of the basic exponential growth function $y=2^x$. Exponential growth functions have a curve that starts near the x-axis for negative x and increases as x increases.

Step2: Determine position relative to x-axis

Since $2^x > 0$ for all real x, multiplying by 3 (positive) keeps $k(x) > 0$ for all x. So the graph is entirely above the x-axis.

Step3: Compare steepness to $y=2^x$

A vertical scaling by a factor greater than 1 makes the function increase more steeply. Since $3>1$, $k(x)$ increases more steeply than $y=2^x$.

Step4: Calculate y for x=0

Substitute $x=0$ into $k(x)$:
$k(0)=3(2^0)=3(1)=3$

Step5: Calculate y for x=1

Substitute $x=1$ into $k(x)$:
$k(1)=3(2^1)=3(2)=6$

Answer:

(a) The graph is entirely above the x-axis and, from left to right, increases more steeply than the graph of $y=2^x$.
(b) For x=0: $(0, 3)$; For x=1: $(1, 6)$