$f(x)$ is a quadratic function with vertex $(1, −2)$, opens up.

Question

I got this on a practice quiz. I couldn't figure it out without at least another point on the graph. $f(x)$ is a quadratic function with vertex $(1, −2)$, opens up.

Answer 1

Equation of the parabola in Vertex Form where the point $(h,k)$ represents the vertex of this parabola is represented as this. $f(x)= a(x-h)^2 + k $

If we take the point you have provided as vertex we will have $h=1$ and $k=-2$ The equation becomes like this=> $f(x)= a(x-1)^2 -2 $

Now what is the role of "opens up" in this problem.

For Parabolas of form $f(x)=ax^2$ there are two cases

If $a>0$ (it is the case where parabola opens up):

For $a<0$(this parabola opens down):

So for the case of having your parabola Open Up ! Choose some $a>0$ to get positive coefficient in your equation $f(x)= a(x-1)^2 -2 $