Secant line slope question I am asked to:
Show the expression for the slope of the secant line through $y=x^2+3x$ at $x=3$ and $x=3+h$ is $msec=9+h$
\begin{align*}
 msec &= \dfrac{f(x+h)-f(x)}{h} \\
 &= \dfrac{(x+h)^2+3(x+h)-(x^2+3x)}{h}\\
 &=\dfrac{(x+h)(x+h)+3x+3h-x^2-3x}{h}\\
 &=\dfrac{2xh+h^2+3h}{h}\\
&=\dfrac{h(2x+h+3)}{h}\\
&=2x+h+3
\end{align*}
The first part seems to work:
\begin{align*}
msec&=2(3)+h+3\\
&=9+h
\end{align*}
Second part: \begin{align*}
msec&=2(3+h)+h+3\\
&=6+2h+h+3\\
&=3h+9
\end{align*}
Am I missing something or is this a mistake / flaw in the question?

You are missing less that you think. On the line where you conclude with $2x+h+3$, you need to stop, and insert $x=3$ to get $9+h$, and then conclude. There is nothing else left to do. There is no second part.
