Question

curve y=f(x),minimum point (1,q),q is a constant...then the quadratic function is 3x(x+p)^2 + 2....find 1.)value p,2.)value q

Answer 1

the post is a bit hard to read, but I think we want

f(x) = 3(x+p)^2 + 2
where the min is at (1,q)

(x+p)^2 has a minimum value of 0 when x = -p. If want x to be 1, then p = -1

So, now we have

f(x) = 3(x-1)^2 + 2
f(1) = 2

So, q=2