Range for the function $f(x) = 3x + 2$ with domain $x > 0$ The function below is defined for continuous domains

Sketch the graph and state the range of the function

Question: $f(x) = 3x + 2$ for the domain $\{x \in \mathbb{R} : x &gt; 0 \}$

The straight line cuts the $y$-axis at $(2, 0)$ but since $x &gt;0$ why is the answer for the range still $f(x) &gt; 2$? Shouldn't you be using $x = 0.1$ or $x = 1$ to find the $y$-intercept since the domain originally stated $x &gt; 0$ (hence $x \not= 0$)

Thanks.

Edit: I have just noticed that the range is $f(x) &gt; 2$ and not $f(x) \ge 2$ which solves everything.
$x&gt;0\Rightarrow 3x+2&gt;2\Rightarrow f(x)\in (2,\infty)$
Yeah... But we first drew the graph of function over $R$ and then smartly erased the part not needed. Your case will yield the same answer.

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