f(x) is a quadratic function such that f(1) = 3 and f(5) = 3. Find the x coordinate of the vertex of the graph of f.
2 answers
help ASAP PLEASEE!
well, since f(1) = f(5), the vertex is at x=3, midway between. So,
y = a(x-3)^2 + k
Now plug in your points to get a and k:
3-k = 4a
k = 3-4a
y = a(x-3)^2 + 3-4a
Unfortunately, since all we have is two points, there's no way to determine exactly where the vertex is. It is somewhere on the line x=3.
However, if we take a=1, then we have
y = (x-3)^2 - 1
y+1 = (x-3)^2
and the vertex is at (3,-1)
y = a(x-3)^2 + k
Now plug in your points to get a and k:
3-k = 4a
k = 3-4a
y = a(x-3)^2 + 3-4a
Unfortunately, since all we have is two points, there's no way to determine exactly where the vertex is. It is somewhere on the line x=3.
However, if we take a=1, then we have
y = (x-3)^2 - 1
y+1 = (x-3)^2
and the vertex is at (3,-1)
3 answers