equation of porabola

Question

equation of a downward opening parabola that has a vertex (-6, 3) and vertical compression of .25

Answer 1

The vertex form of a quadratic equation (parabola) is as follows

y = a ( x - h)² + k

Where a is the stretch/compressions, (h,k) is the vertex.

Thus if the vertex is at (-6,3) then h=-6 and k=3, and a=.25

So combining that into the vertex form of the equation we get

y=.25(x+6)²+3

Now, if the answer must be put into standard form you will need to rearrange the equation into the form bellow

y=ax²+bx+c

To do that you would follow these steps

1. y=.25(x+6)²+3 you would need to perform the square first
2. y=.25(x²+12x+36)+3 now you will need to distribute the .25
3. y=.25x²+3x+9+3 and finally combine like terms for the constants 9 and 3
4. y=.25x²+3x+12