Question

the graph of $f(x) = 3^x$ is reflected across the $x$-axis, shifted downward 7 units, and then shifted left 2 units. let $g(x)$ represent the new function. what is an equation for this function? $g(x) = $ help (formulas) what is the $y$-intercept of $g(x)$? enter the answer as an ordered pair. help (points) what is the domain of $g(x)$? enter the answer as an interval. help (intervals) what is the range of $g(x)$? enter the answer as an interval. help (intervals)

Explanation:

Step1: Apply transformation rules

For a function \(y = f(x)\), reflecting across the \(x -\)axis gives \(y=-f(x)\), shifting left \(2\) units gives \(y =-f(x + 2)\), and shifting down \(7\) units gives \(y=-f(x + 2)-7\). Given \(f(x)=3^{x}\), then \(g(x)=-3^{x + 2}-7\).

Step2: Find the \(y -\)intercept

The \(y -\)intercept occurs when \(x = 0\). Substitute \(x = 0\) into \(g(x)\): \(g(0)=-3^{0+2}-7=-3^{2}-7=-9 - 7=-16\). So the ordered pair is \((0,-16)\).

Step3: Determine the domain

The domain of an exponential function of the form \(y = a\cdot b^{x + h}+k\) (where \(a=-1\), \(b = 3\), \(h = 2\), \(k=-7\)) is all real numbers. In interval notation, the domain is \((-\infty,\infty)\).

Step4: Determine the range

Since \(3^{x+2}>0\), then \(-3^{x + 2}<0\), and \(-3^{x+2}-7<-7\). In interval notation, the range is \((-\infty,-7)\).

Answer:

\(g(x)=-3^{x + 2}-7\); \(y -\)intercept: \((0,-16)\); Domain: \((-\infty,\infty)\); Range: \((-\infty,-7)\)