the range of which function is $(2,\\infty)$?\n$y = 2^x$\n$y = 2(5^x)$\n$y = 5^{x + 2}$\n$y = 5^x+2$

Answer:

D. $y = 5^{x}+2$

Explanation:

Step1: Recall exponential - function properties

The general form of an exponential function is $y = a\cdot b^{x}+k$, where $b>0,b\neq1$. The range of $y = b^{x}$ is $(0,\infty)$ for $b > 0,b\neq1$.

Step2: Analyze option A

For $y = 2^{x}$, the range is $(0,\infty)$ since the general form of the exponential function $y = b^{x}$ with $b = 2>0$ has a range of all positive real - numbers.

Step3: Analyze option B

For $y = 2(5^{x})$, since $5^{x}>0$ for all real $x$, then $y = 2(5^{x})>0$. The range is $(0,\infty)$.

Step4: Analyze option C

For $y = 5^{x + 2}=5^{x}\cdot5^{2}=25\cdot5^{x}$, and since $5^{x}>0$ for all real $x$, then $y = 25\cdot5^{x}>0$. The range is $(0,\infty)$.

Step5: Analyze option D

For $y = 5^{x}+2$, we know that the range of $y = 5^{x}$ is $(0,\infty)$. When we add 2 to $5^{x}$, we shift the graph of $y = 5^{x}$ vertically upwards by 2 units. So the range of $y = 5^{x}+2$ is $(2,\infty)$.