I am not sure how to begin solving this.
With two variables $p$ and $q$ that need to be solved for in order for all three to be continuous.
$$ f(x) = \left\{ \begin{array}{rl} x+p &\mbox{ if $x \le 0$} \\ x^2+q &\mbox{if $0 < x < 5$}\\ 6x+5 &\mbox{if $x \ge 5$} \end{array} \right. $$
My intuition to start is that I need to check the functions at the points $x =0$ and $x=5$
For example $6(5)+5 = 5^2+q$ so then $q = 10$
For example $(0) + p = 0^2 + 10$ so then $p=10$. That doesn't make sense....
Any help to get me going would be much appreciated.
Since $f(x)$ is a continuous function, so it is also continuous at $x=0$ and $x=5$.
So $$\lim_{x\to 0^+} f(x)=f(0)=\lim_{x\to 0^-} f(x)$$and$$\lim_{x\to 5^+} f(x)=f(5)=\lim_{x\to 5^-} f(x)$$
Now $$\lim_{x\to 0^+} f(x)=f(0)=\lim_{x\to 0^-} f(x)\implies \lim_{x\to 0^+} (x^2 + q)=p=\lim_{x\to 0^-} (x+p)\implies q=p$$
Again $$\lim_{x\to 5^+} f(x)=f(5)=\lim_{x\to 5^-} f(x)\implies \lim_{x\to 5^+} (6x+5)=6.5+5=\lim_{x\to 5^-} (x^2 + q)\implies 35=35=25+q \implies q=10$$
Hence $p=q=10$