Let the function $p$ be defined as $p(x)=\dfrac{(x-c)^2+12}{2c}$, where $c$ is a constant. If $p(c)=2$, what is the value of $p(4)$ ?

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We first solve for $c$ by using the given information $p(c)=2$.

$$p(c)=2$$ $$\frac{(c-c)^2+12}{2c}=2$$ $$\frac{0^2+12}{2c}=2$$ $$\frac{12}{2c}=2$$ $$c=3$$

We can $c=3$ into function $p$ so that:

$$p(x)=\frac{(x-c)^2+12}{2c}$$ $$p(x)=\frac{(x-3)^2+12}{6}$$

Finding $p(4)$

$$p(4)=\frac{(4-3)^2+12}{6}$$ $$= \frac{1+12}{6}$$ $$= \frac{13}{6}$$