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 above the x - axis and from left to right, falls rises steeply than the graph of y = 2^x.

Explanation:

Step1: Analyze the exponential function

The function $k(x)=3(2^x)$ is an exponential function with base $2>1$, and vertical stretch factor 3.

Step2: Describe shape for (a)

For $b>1$, $b^x>0$ for all $x$, so $3(2^x)>0$. As $x$ increases, $2^x$ increases, so $k(x)$ rises. The vertical stretch means it rises at the same rate of steepness relative to $y=2^x$ but scaled vertically.

Step3: Calculate point for $x=0$

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

Step4: Calculate point 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 rises from left to right, equally steeply than the graph of $y=2^x$.
(b) For $x=0$, the second coordinate is 3; for $x=1$, the second coordinate is 6.
Or formatted as points: $(0, 3)$ and $(1, 6)$