Question

I need a bit of help... please

determine the quadratic function if the vertex is (1,-4) and the y intercept is -3

answer must be in the form f(x)=ax^2+bx+c

Answer 1

If the vertex is (1,-4), then (h,k) = (1,-4) meaning that h = 1 and k = -4


y = a(x-h)^2 + k


y = a(x-1)^2 + (-4)


y = a(x-1)^2 - 4


We're given that it has a y-intercept of -3, so x = 0 and y = -3. Plug these additional values in


y = a(x-1)^2 - 4


-3 = a(0-1)^2 - 4


-3 = a(-1)^2 - 4


-3 = a(1) - 4


-3 = a - 4


-3+4 = a


1 = a


a = 1


So the function in vertex form is f(x) = 1(x-1)^2 - 4 which can be written as f(x) = (x-1)^2 - 4


Now convert to standard form


f(x) = (x-1)^2 - 4


f(x) = x^2-2x+1 - 4


f(x) = x^2-2x-3


So the answer is f(x) = x^2-2x-3

Answer 2

ahh i see ;)

Answer 3

that's great :)