\(y'(t)=mt
\\
\\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C\)
Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.
\(y=at^2+b
\\
\\3=a\times 2^2-2
\\
\\3=4a-2
\\
\\5=4a
\\
\\a= \frac{5}{4}
\\
\\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C
\\ \\ \frac{m}{2} = \frac{5}{4}
\\
\\m= \frac{5}{2}
\\
\\C=-2
\\
\\y(t_0)=-4
\\
\\-4= \frac{5}{4} t_0^2-2
\\
\\ -2= \frac{5}{4} t_0^2
\\
\\ t_0^2=- \frac{8}{5}
\\
\\t_0=\pm \sqrt{- \frac{8}{5} }\)
