### Instruction: Given that $(-2,3)$ is on the graph of $y=f(x)$, find a point on the graph of $y=f(2x+1)+3$. Express your answer as an ordered pair $(a,b)$ where $a$ and $b$ are real numbers.### Response: I need to find a point on the graph of a transformed version of $f(x)$.
I notice that the new function has a horizontal stretch by a factor of $\frac{1}{2}$, a horizontal shift to the left by $\frac{1}{2}$, and a vertical shift up by $3$.
This means that if I take a point $(x,y)$ on the original graph of $f(x)$, I can apply these transformations to get a corresponding point $(x',y')$ on the new graph of $f(2x+1)+3$.
Specifically, $x'=\frac{1}{2}x-\frac{1}{2}$ and $y'=y+3$.
Since I know that $(-2,3)$ is on the graph of $f(x)$, I can plug in these values to get a point on the new graph.
$x'=\frac{1}{2}(-2)-\frac{1}{2}=-1-0.5=-1.5$ and $y'=3+3=6$.
Therefore, a point on the graph of $y=f(2x+1)+3$ is $(-1.5,6)$.