The graph of a quadratic function f(x) is shown above. It has a vertex at (-2,4) and passes the point (0,2). Find the quadratic function
y=a(x-(-2))^2+4
y=a(x+2)+4
2=a(x+2)+4
-4 -4
-2=a(0+2)/2
-1+ a
function -1(x+2)+4 ?
Im not sure where Im going wrong but my answer was incorrect. Please help
from the given vertex, the equation must be
y = a(x+2)^2 + 4
but (0,2) lies on it, so
2 = a(0+2)^2 + 4
2 = 4a + 4
4a=-2
a = -1/2
y = (-1/2)(x+2)^2 + 4
To find the quadratic function, we can use the given information about the vertex and a point that lies on the graph of the function.
The vertex form of a quadratic function is given by:
f(x) = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola. In this case, we are given that the vertex is (-2, 4), so h = -2 and k = 4.
Using the given point (0, 2), we can substitute these values into the equation to find the value of a:
2 = a(0 - (-2))^2 + 4
2 = a(2)^2 + 4
2 = 4a + 4
2 - 4 = 4a
-2 = 4a
a = -2/4
a = -1/2
Now that we have the value of a, we can substitute it back into the vertex form equation:
f(x) = (-1/2)(x - (-2))^2 + 4
Simplifying further:
f(x) = (-1/2)(x + 2)^2 + 4
So, the quadratic function is f(x) = (-1/2)(x + 2)^2 + 4.
The graph of a quadratic function f(x) is shown above. It has a vertex at (2,4) and passes the point (0,2). Find the quadratic function.
F(x)=