SOLUTION: The graph of an parabola opens up, as a vertex at (-1,1) and an intersection at (0,3). Write the equation of the parabola in y=ax^2+bx+c format. None of the answers they give make

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The graph of an parabola opens up, as a vertex at (-1,1) and an intersection at (0,3). Write the equation of the parabola in y=ax^2+bx+c format. None of the answers they give make       Log On


   

Question 1028757: The graph of an parabola opens up, as a vertex at (-1,1) and an intersection at (0,3). Write the equation of the parabola in y=ax^2+bx+c format. None of the answers they give make sense. Any help will be appreciated.
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
The "parent equation" for simple parabolas is
y=x%5E2

Using vertex form

y-k=a%28x-h%29%5E2 where h,k is the vertex, we have:

y-1=a%28x%2B1%29%5E2

now, we have a standard parabola, shifted to get your vertex. But, if you enter 0 for X, you get 2, not the intersection 3 you need. This is where 'a' comes in. It will leave the vertex alone, but transform the shape to 'pull up' the parabola. We want (0,3) to be a solution.
3-1=a%280%2B1%29%5E2

2=a%281%29%5E2 so a=2

y-1=2%28x%2B1%29%5E2

and you rearrange to

y-1=2x%5E2%2B2x%2B1

y-1=2x%5E2%2B4x%2B2

y=2x%5E2%2B4x%2B3