determine the qudratic function whose vertex is (2,-4) and passes through
(3,-1)
fx=
since we know the vertex, let it be
y = a(x-2)^2 - 4
but (3,-1) lies on it
-1 = a(1) - 4
a = 3
y = 3(x-2)^2 - 4
new question using the square to descibe the graph of fx= -8 -16x-x^2
To determine the quadratic function fx, we need to find the equation in the form "y = ax^2 + bx + c" by using the given information.
1. The vertex form of a quadratic function is "y = a(x - h)^2 + k", where (h, k) represents the vertex.
In this case, the vertex is (2,-4), so our equation becomes: y = a(x - 2)^2 - 4.
2. We also know the quadratic function passes through the point (3,-1).
Substitute the coordinates (x, y) = (3,-1) into the equation and solve for the coefficient 'a':
-1 = a(3 - 2)^2 - 4
-1 = a(1)^2 - 4
-1 = a - 4
a = 3 (by adding 4 to both sides)
3. Now, we have the value of 'a' as 3. Substitute it back into the vertex form equation:
y = 3(x - 2)^2 - 4
Therefore, the quadratic function with a vertex at (2,-4) and passing through (3,-1) is:
fx = 3(x - 2)^2 - 4