SOLUTION: Determine the quadratic function f whose vertex is (1,-3) and passes through (2,-1). f(x) = (Give the answer in the form f(x) = ax^2 + bx + c.) This is very confusing to me.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Determine the quadratic function f whose vertex is (1,-3) and passes through (2,-1). f(x) = (Give the answer in the form f(x) = ax^2 + bx + c.) This is very confusing to me.      Log On


   



Question 669342: Determine the quadratic function f whose vertex is (1,-3) and passes through (2,-1).
f(x) =
(Give the answer in the form f(x) = ax^2 + bx + c.)
This is very confusing to me.
I know that the vertex (h,k) = (1,-3) and that this is a parabola that passes through the point of (2,-1)(x being 2 and y being -1) but I am not sure how to do what they are asking me here. Please Help!!
Nichole

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the quadratic function f whose vertex is (1,-3) and passes through (2,-1).
f(x) =
(Give the answer in the form f(x) = ax^2 + bx + c.)
**
Using standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex
plug in given coordinates of vertex (1,-3)
y=A(x-1)^2-3
plug in given coordinates of point: (2,-1), then solve for A
-1=A(2-1)^2-3
-1=A-3
A=2
Equation of parabola: (quadratic function):
F(x)=2(x-1)^2-3
F(x)=2(x^2-2x+1)-3
F(x)=2x^2-4x+2-3
F(x)=2x^2-4x-1
see graph below:
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+2x%5E2-4x-1%29+