Question

which of the following shows the graph of $y = -(2)^x - 1$?

Explanation:

Step1: Analyze the parent function

The parent function is \(y = 2^{x}\), which is an exponential growth function passing through \((0,1)\) and approaching \(y = 0\) as \(x\to-\infty\).

Step2: Consider the transformations

• The negative sign in \(y=-(2)^{x}-1\) reflects the parent function \(y = 2^{x}\) over the \(x\) - axis. So the function \(y=-2^{x}\) has a \(y\) - intercept at \((0, - 1)\) (since when \(x = 0\), \(y=-2^{0}=-1\)) and approaches \(y = 0\) as \(x\to-\infty\) (but from below the \(x\) - axis).
• The \(-1\) in \(y=-(2)^{x}-1\) shifts the graph of \(y =-2^{x}\) down by 1 unit. So when \(x = 0\), \(y=-(2)^{0}-1=-1 - 1=-2\). As \(x\to-\infty\), \(y =-(2)^{x}-1\) approaches \(y=-1\) (because \(\lim_{x\to-\infty}(2^{x}) = 0\), then \(\lim_{x\to-\infty}(-2^{x}-1)=-1\)). As \(x\to+\infty\), \((2)^{x}\to+\infty\) and \(y=-(2)^{x}-1\to-\infty\).

Answer:

The graph that has a \(y\) - intercept at \((0,-2)\), approaches \(y = - 1\) as \(x\to-\infty\) and goes to \(-\infty\) as \(x\to+\infty\) is the correct graph of \(y=-(2)^{x}-1\).