Find the quadratic function

Question

Find the quadratic function f(x)= ax^2+bx+c that goes through (2, 0) and has a local minimum at (0, -1)

Answer 1

(0,-1) minimum= vertex

quadratic through (2,0)

y=a(x^2)-1 in vertex form

plug in the point to solve for a

0=4a-1

a= 1/4

y =(1/4)x^2-1

f(x) = (x^2)/4 -1

Answer 2

Hey there, so since we know that this is a quadratic we know that the local minimum is the vertex, so we can start by setting up the vertex form of a quadratic like this:

y=a(x-0)^2-1

y=ax^2-1

To find a we just simply have to plug the point it goes through into the equation and solve for a.

0=a2^2-1

0=4a-1

1=4a

a=1/4

Then we plug it into the formula.

f(x)=1/4x^2-1

Let me know if you have any questions