The vertex of a parabola represented by f(x)=x^2-4x+3x has coordinates (2, -1). Find the Coordninates of the vertex of the parabola defined by g(x)=f(x-2).
Hi so I just have no idea what to do can someone help? Thank you
To find the vertex of the parabola defined by g(x)=f(x-2), we need to first determine the equation of g(x).
Substituting x-2 for x in the equation of f(x), we get:
g(x) = (x-2)^2 - 4(x-2) + 3
Simplifying this, we get:
g(x) = x^2 - 8x + 17
Now we can find the vertex of g(x).
The x-coordinate of the vertex is given by:
x = -b/2a
where a and b are the coefficients of x^2 and x, respectively, in the equation of g(x).
In this case, a = 1 and b = -8, so:
x = -(-8)/(2*1) = 4
The y-coordinate of the vertex is given by:
y = g(4) = 4^2 - 8(4) + 17 = -3
Therefore, the coordinates of the vertex of the parabola defined by g(x)=f(x-2) are (4, -3).