#### Answer

$\color{blue}{f(x)=x^2-\frac{14}{3}x+\frac{8}{3}}$

#### Work Step by Step

RECALL:
The zeros of the quadratic function $f(x) =(x+a)(x+b)=0$ are $x=-a$ and $x=-b$.
The given quadratic function has the zeros $x=\frac{2}{3}$ and $x=4$.
Using the rule mentioned in the recall part above, then the function is:
$f(x)=(x-\frac{2}{3})(x-4)
\\f(x)=x(x)+x(-4) - \frac{2}{3}(x) - \frac{2}{3}(-4)
\\f(x)=x^2-4x-\frac{2}{3}x+\frac{8}{3}
\\f(x)=x^2-\frac{12}{3}x-\frac{2}{3}x+\frac{8}{3}
\\\color{blue}{f(x)=x^2-\frac{14}{3}x+\frac{8}{3}}$