Question

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

Answer 1

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

Answer 2

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)=

Answer 3

The vertex of the parabola defined byf(x)=−5x2+4x+1is the point (, ).