Write a quadratic equation that goes through the points (2,3) (3,2) and (4,3) ?
You could let \(f(x)=ax^2+bx+c\) and solve the system, but that is messy. Alternatively, we can plot the points. This shows us that the vertex of the parabola is simply (3,2)! So our graph is shifted three to the right, and 2 up. This means that \(f(x)=(x-3)^2+2=x^2-6x+11\).
Graph --> https://www.desmos.com/calculator/xvkbk8bhpl