Question

1. use the x - y table to graph the function. then find the characteristics below.

$f(x)=-2^{x}+5$

• growth or decay?
• transformations?
• asymptote at
• domain:
• range:
• y - intercept:

Explanation:

Step1: Analyze the base of the exponential function

For the function \(y = a\cdot b^{x}+k\), in \(f(x)=-2^{x}+5\), the base \(b = 2>1\). But since there is a negative sign in front of \(2^{x}\), as \(x\) increases, \(2^{x}\) increases and \(-2^{x}\) decreases. So it is a decay - related function (because of the negative coefficient of the exponential term).

Step2: Identify transformations

The parent function is \(y = 2^{x}\). The transformation from \(y = 2^{x}\) to \(y=-2^{x}+5\) involves a reflection over the \(x\) - axis (due to the negative sign) and a vertical shift up by \(5\) units (due to the \(+5\)).

Step3: Find the horizontal asymptote

For an exponential function of the form \(y = a\cdot b^{x}+k\), the horizontal asymptote is \(y = k\). In \(f(x)=-2^{x}+5\), \(k = 5\), so the horizontal asymptote is \(y = 5\).

Step4: Determine the domain

For any exponential function \(y=a\cdot b^{x}+k\), the domain is all real numbers. Because we can substitute any real - valued \(x\) into the function \(f(x)=-2^{x}+5\). So the domain is \((-\infty,\infty)\) (or \(\mathbb{R}\)).

Step5: Determine the range

Since \(2^{x}>0\) for all \(x\in\mathbb{R}\), then \(-2^{x}<0\). Adding \(5\) to both sides of the inequality \(-2^{x}<0\), we get \(-2^{x}+5 < 5\). So the range is \((-\infty,5)\).

Step6: Find the y - intercept

To find the \(y\) - intercept, set \(x = 0\). Then \(f(0)=-2^{0}+5\). Since \(2^{0}=1\), \(f(0)=-1 + 5=4\). So the \(y\) - intercept is \((0,4)\).

Answer:

• growth or decay? Decay
• transformations? Reflection over the \(x\) - axis and vertical shift up \(5\) units
• asymptote at \(y = 5\)
• domain: \((-\infty,\infty)\)
• range: \((-\infty,5)\)
• y - intercept: \((0,4)\)