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 \( p(4) \)

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