Answer
$$K = \frac{4}{{17\sqrt {17} }},{\text{ }}r = \frac{{17\sqrt {17} }}{4}$$
Work Step by Step
$$\eqalign{
& y = 2{x^2} + 3,{\text{ }}x = - 1 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^2} + 3} \right] \cr
& y'\left( x \right) = 4x \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {4x} \right] \cr
& y''\left( x \right) = 4 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| 4 \right|}}{{{{\left( {1 + {{\left[ {4x} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Evaluate at }}x = - 1 \cr
& K = \frac{{\left| 4 \right|}}{{{{\left( {1 + {{\left[ {4\left( { - 1} \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{4}{{17\sqrt {17} }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{17\sqrt {17} }}{4} \cr} $$
