Suppose that $G(x)=\log _{2}(2 x+2)-3$
(a) What is the domain of G?
(b) What is $G(1)$ ? What point is on the graph of $G$ ?
(c) If $G(x)=2$, what is $x$ ? What point is on the graph of $G$ ?
(d) What is the zero of G?
(a) The domain of $\mathrm{G}$ is $(-1, \infty)$. (Type your answer in interval notation.)
(b) $G(1)=-1$
The point $(1,-1)$ is on the graph of $G$. (Type an ordered pair.)
(c) If $G(x)=2$, then $x=15$
The point $\square$ is on the graph of $\mathrm{G}$. (Type an ordered pair.)

Step 1 ：For the given function to have a real value, the argument of the logarithm, i.e., \(2x+2\), must be greater than 0. So, we have \(2x+2>0\).

Step 2 ：Solving the inequality, we subtract 2 from both sides to get \(2x>-2\).

Step 3 ：Then, we divide both sides by 2 to get \(x>-1\).

Step 4 ：So, the domain of \(G(x)\) is \(x \in \boxed{(-1, \infty)}\) in interval notation.

Step 5 ：To find \(G(1)\), we substitute \(x=1\) into the function to get \(G(1)=\log _{2}(2*1+2)-3=\log _{2}(4)-3=2-3=-1\).

Step 6 ：So, the point \((1,-1)\) is on the graph of \(G\).

Step 7 ：If \(G(x)=2\), we set the function equal to 2 and solve for \(x\): \(\log _{2}(2x+2)-3=2\).

Step 8 ：Adding 3 to both sides gives \(\log _{2}(2x+2)=5\).

Step 9 ：Using the property of logarithms that \(a=\log _{b}(c)\) is equivalent to \(b^a=c\), we rewrite the equation as \(2^5=2x+2\) or \(32=2x+2\).

Step 10 ：Subtracting 2 from both sides gives \(30=2x\), and dividing both sides by 2 gives \(x=15\).

Step 11 ：So, the point \((15,2)\) is on the graph of \(G\).

Step 12 ：The zero of \(G\) is the value of \(x\) that makes \(G(x)=0\). We set the function equal to 0 and solve for \(x\): \(\log _{2}(2x+2)-3=0\).

Step 13 ：Adding 3 to both sides gives \(\log _{2}(2x+2)=3\).

Step 14 ：Using the property of logarithms that \(a=\log _{b}(c)\) is equivalent to \(b^a=c\), we rewrite the equation as \(2^3=2x+2\) or \(8=2x+2\).

Step 15 ：Subtracting 2 from both sides gives \(6=2x\), and dividing both sides by 2 gives \(x=3\).

Step 16 ：So, the zero of \(G\) is \(x=3\).