Answer
Graph
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{{{x^2} + 16}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = \frac{{{{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^2} + 16}} \to y = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = \frac{{{x^2}}}{{{x^2} + 16}} \cr
& x = 0 \cr
& x{\text{ - intercept }}\left( {0,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} + 16}}} \right] \cr
& y' = \frac{{\left( {{x^2} + 16} \right)\left( {2x} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} + 16} \right)}^2}}} \cr
& y' = \frac{{2{x^3} + 32x - 2{x^3}}}{{{{\left( {{x^2} + 16} \right)}^2}}} \cr
& y' = \frac{{32x}}{{{{\left( {{x^2} + 16} \right)}^2}}} \cr
& {\text{Let }}y' = 0 \cr
& 32x = 0 \cr
& x = 0 \cr
& {\text{Evaluate }}y'\left( { - 1} \right){\text{ and }}y'\left( 1 \right) \cr
& y\left( { - 1} \right)' = \frac{{32\left( { - 1} \right)}}{{{{\left( {{{\left( { - 1} \right)}^2} + 16} \right)}^2}}} < 0 \cr
& y\left( 1 \right)' = \frac{{32\left( 1 \right)}}{{{{\left( {{{\left( 1 \right)}^2} + 16} \right)}^2}}} > 0 \cr
& y\left( { - 1} \right)' < 0,{\text{ }}y\left( 0 \right)' = 0,{\text{ }}y\left( 1 \right)' > 0 \cr
& {\text{There is a relative minimum at }}x = 0 \cr
& y\left( 0 \right) = \frac{{{{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^2} + 16}} = 0 \cr
& {\text{Relative minumum at }}\left( {0,0} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& \frac{{{x^2}}}{{{x^2} + 16}} \cr
& {x^2} + 16 = 0,{\text{ No real solutions}} \cr
& {\text{There are no vertical asymptotes}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{x^2} + 16}} = 1 \cr
& \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}}}{{{x^2} + 16}} = 1 \cr
& {\text{Horizontal asymptote }}y = 1 \cr
& \cr
& {\text{*Symmetry}} \cr
& f\left( { - x} \right) = \frac{{{{\left( { - x} \right)}^2}}}{{{{\left( { - x} \right)}^2} + 16}} = \frac{{{x^2}}}{{{x^2} + 16}} \cr
& f\left( { - x} \right) = f\left( x \right){\text{ The function is even}} \cr
& {\text{Symmetry about the }}y{\text{ - axis}} \cr
& \cr
& {\text{Graph}} \cr} $$
