Find $f$ such that $f^{\prime }(x)=8x^2+5x-2$ and $f(0)=2$

Step 1 ：We are given that the derivative of the function $f$ is $f^{\prime }(x)=8x^2+5x-2$ and that the function passes through the point $(0,2)$.

Step 2 ：To find the function $f$, we can integrate the derivative to find the antiderivative.

Step 3 ：The antiderivative of $f^{\prime }(x)=8x^2+5x-2$ is $f(x)=\frac{8}{3}{x}^3+\frac{5}{2}{x}^2-2x+C$, where $C$ is the constant of integration.

Step 4 ：We can determine the value of $C$ by using the point $(0,2)$, which the function $f$ passes through.

Step 5 ：Substituting $x=0$ and $f(0)=2$ into the equation gives $C=2$.

Step 6 ：Substituting $C=2$ back into the equation gives the function $f(x)=\frac{8}{3}{x}^3+\frac{5}{2}{x}^2-2x+2$.

Step 7 ：$\boxed{{\displaystyle f(x)=\frac{8}{3}{x}^3+\frac{5}{2}{x}^2-2x+2}}$ is the function such that $f^{\prime }(x)=8x^2+5x-2$ and $f(0)=2$.