Answer
$f(x)=4(x-2)^2-3$ or $f(x)=4x^2-16x+13$.
Work Step by Step
To find a quadratic function $f$ that has the assigned properties, first we need to write down a quadratic function's standard form:
$f(x)=a(x-h)^2+k$
As we know, the coordinates of the vertex describe the minimum/maximum value of that quadratic function and taking the form of $(h,k)$. Therefore, we can now rewrite our function as:
$f(x)=a(x-2)^2-3$
In this question, they also give us the position of another point that belongs to the parabola. We can substitute the value of the coordinates to the function to find $a$.
$f(x)=a(x-2)^2-3$
$1=a(3-2)^2-3$
$1=a*1-3$
$a-3=1$
$a=4$
In conclusion, the desired function is $f(x)=4(x-2)^2-3$ or $f(x)=4x^2-16x+13$.
