Let f(x)=3x^2-6x+k, where k is a constant. If the y-coordinate of the vertex of the graph of y=f(x) is 7, then k=?
f(x) = 3x^2-6x+k
= 3(x^2-2x) + k
= 3(x^2-2x+1) + k-3
= 3(x-1)^2 + k-3
The vertex is at (1,k-3), so
k-3 = 7
k = 10
= 3(x^2-2x) + k
= 3(x^2-2x+1) + k-3
= 3(x-1)^2 + k-3
The vertex is at (1,k-3), so
k-3 = 7
k = 10