The point $P(1,\frac{1}{4})$ lies on the curve $y=\frac{x}{3+x}.($Round your answers to within three decimal places if necessary, but do not round until your final computation.$)$

a$.$ If $Q$ is the point $(x,\frac{x}{3+x}),$ find the slope of the secant line $PQ$ for the following values of $x.$

$\backslash$table$[[x,$Slope,$x,$Slope$],[0\mathrm{.}5,,1\mathrm{.}5,],[0\mathrm{.}9,,1\mathrm{.}1,],[0\mathrm{.}99,,1\mathrm{.}01,],[0\mathrm{.}999,,1\mathrm{.}001,]]$

b$.$ Guess the slope of the tangent line to the curve at $P.$

Slope: $$

c$.$ Using the slope from Part b$.,$ find the equation of the tangent line to the curve at $P.$

Tangent line: $y=$