Question

In a quadratic Function the standard formula is f(x)=a(x-h)^2+k
so the vertex is (h,k)
my question is if i get (x+h)^2 does that make h negative when I find the vertex?
and If its -(x+h)^2 does that change the sign of h?

Answer 1

Ok this is what you need to know. What you have there is the quadratic formula given in vertex form. You can also obtain that form by completing the square but we'll leave that for now. Anyway to move on:\[\bf f(x)=a(x-h)^2+k\]This implies that the vertex is at \(\bf (h,k)\) and is unaffected by the value of \(\bf a\), it doesn't matter if \(bf a\) is positive or negative, it doesn't affect the vertex.

Now if you get \(\bf-(x+h)^2+k\) then re-write it as:\[\bf -(x-(-h))^2+k\]This now implies that the vertex is at \(\bf (-h,k)\). So basically the sign changed of 'h'. @eragon4

Answer 2

Thank so much!!