SOLUTION: Hi there, I have to find the quadratic equation that has vertex (-1, -3) and y-intercept (0, 1). I plugged the vertex into the vertex form and got {{{a(x + 1)^2 - 3}}} The

Algebra ->  Equations -> SOLUTION: Hi there, I have to find the quadratic equation that has vertex (-1, -3) and y-intercept (0, 1). I plugged the vertex into the vertex form and got {{{a(x + 1)^2 - 3}}} The      Log On


   

Question 367166: Hi there, I have to find the quadratic equation that has vertex (-1, -3) and y-intercept (0, 1).
I plugged the vertex into the vertex form and got
a%28x+%2B+1%29%5E2+-+3
Then I plugged in the given y-intercept and solved for a
-1+=+a%280+%2B+1%29%5E2+-+3
a+=+2
So my vertex form will be 2%28x%2B1%29%5E2-3. Converting this to general form...
2%28x%5E2+%2B+2x+%2B+1%29+-+3
%282x%5E2+%2B+4x+%2B+2%29+-+3
2x%5E2+%2B+4x+-+1
Did I do this correctly? I've never seen a problem like this and couldn't find an example in the book. Thanks for your time and have a good day :)
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi, 

Proceeded exactly as one should.

If you have used the correct value for the y-intercept. all's well.

**Listed is a y-intercept (0,1)

1+=+a%280+%2B+1%29%5E2+-+3

a = 4

vertex form

y = 4*(x+1)^2 - 3

general form

4x^2 + 8x + 1