Question

below is the graph of $y = e^{x}$.transform it to make the graph of $y=-e^{x - 1}$.give the range and domain of $y=-e^{x - 1}$ using interval notation.

Explanation:

Step1: Analyze the domain

For any exponential function of the form \(y = a\cdot e^{bx + c}+d\), the domain is all real numbers. There is no restriction on the value of \(x\) for the function \(y=-e^{x}-1\) since we can substitute any real - number value of \(x\) into the expression \(e^{x}\).

Step2: Analyze the range

We know that the range of the basic exponential function \(y = e^{x}\) is \((0,\infty)\).
For the function \(y=-e^{x}-1\), first consider the transformation of \(y = e^{x}\) to \(y=-e^{x}\). When we multiply \(e^{x}\) by \(- 1\), we reflect the graph of \(y = e^{x}\) over the \(x\) - axis. The range of \(y=-e^{x}\) is \((-\infty,0)\) because if \(y = e^{x}>0\), then \(y=-e^{x}<0\).
Then, for the function \(y=-e^{x}-1\), we shift the graph of \(y=-e^{x}\) down by 1 unit. Using the property of vertical shifts, if \(y = f(x)\) has a range of \(R_f\), then \(y=f(x)-k\) has a range of \(R_f - k\).
So, if the range of \(y=-e^{x}\) is \((-\infty,0)\), then the range of \(y=-e^{x}-1\) is \((-\infty,- 1)\)

Answer:

• Domain: \((-\infty,\infty)\)
• Range: \((-\infty,-1)\)