Answer
$f(x)=\frac{1}{4}(x-2)^2-1$
Work Step by Step
Step 1. Assume the quadratic has a standard form $f(x)=a(x-h)^2+k$
Step 2. Given vertex at $(2,-1)$, we have $h=2, k=-1$, thus $f(x)=a(x-2)^2-1$
Step 3. As $(0,0)$ is on the parabola, we have $0=a(0-2)^2-1$ which gives $a=\frac{1}{4}$
Step 4. We conclude that the function is $f(x)=\frac{1}{4}(x-2)^2-1$
