The graph of a quadratic function f(x) is shown above. It has a vertex at (−2,1) and passes the point (0,−1). Find the quadratic functi

Question

The graph of a quadratic function  is shown above. It has a vertex at  (-2,1)and passes the point(0-1) . Find the quadratic function

F(x)=

Answer 1

-b/(2a) = -2

-b = -4a

b = 4a

f(x) = ax^2 + bx + c

= ax^2 +(4a)x + c

f(-2) = 1 --> 1 = a(-2)^2 + (4a)(-2) + c

= 4a + - 8a + c

= c - 4a

f(0,-1) --> -1 = a(0)^2 + (4a)(0) + c

= c

So, per the y-intercept (0,1) --> c = -1

then 1 = -1 - 4a

2 = -4a

so a = -1/2 ---> b = 4a = 4(-1/2) = -2

the quadratic function is:

f(x) = (-1/2)x^2 - 2x - 1

Note that f(0) = -1, f(-2) = (-1/2)(4) - 2(-2) - 1 = -2+4-1= 2

and, by the completing of the square...

f(x)=

(-1/2) ( x^2 + 4x ) - 1

(-1/2) (x^2 + 4x + 4) - 1 +2 <-- (-1/2)*4 = -2 was added inside the parenthesis!!!

(-1/2) (x + 2)^2 + 1

so the vertex is (-2,1)

The official answer is f(x) = (-1/2)x^2 - 2x - 1