Question

the graph of ( f(x)=2^{x} ) has which of the following features?
a. a constant first difference
b. symmetry about the origin
c. a vertex
d. a horizontal asymptote at ( y = 0 )

Explanation:

Step1: Analyze option a

The function \(y = 2^{x}\) is an exponential function. The first - difference of \(y = 2^{x}\) is \(\Delta y=2^{x + 1}-2^{x}=2^{x}(2 - 1)=2^{x}\), which is not a constant.

Step2: Analyze option b

For a function to be symmetric about the origin, \(f(-x)=-f(x)\). For \(y = 2^{x}\), \(f(-x)=2^{-x}=\frac{1}{2^{x}}\) and \(-f(x)=-2^{x}\). Since \(2^{-x}
eq - 2^{x}\) for all \(x\), the function \(y = 2^{x}\) is not symmetric about the origin.

Step3: Analyze option c

The function \(y = 2^{x}\) is an exponential function of the form \(y = a^{x}(a>1)\). It has no vertex. The general form of a function with a vertex is a quadratic function \(y = ax^{2}+bx + c(a
eq0)\) which has a vertex at \(x =-\frac{b}{2a}\).

Step4: Analyze option d

We know that \(\lim_{x
ightarrow-\infty}2^{x}=0\). The horizontal asymptote of a function \(y = f(x)\) is found by evaluating \(\lim_{x
ightarrow\pm\infty}f(x)\). For \(y = 2^{x}\), as \(x
ightarrow-\infty\), \(y = 2^{x}\) approaches \(0\). So, \(y = 0\) is a horizontal asymptote.

Answer:

d. A horizontal asymptote at \(y = 0\)