Answer is D... the -3 outside the absolute/parenthesis shifts is down
State the vertex and axis of symmetry of the graph of y=ax^2+c
General form of quadratic equation is ![y=ax^2 + bx +c](https://tex.z-dn.net/?f=y%3Dax%5E2%20%2B%20bx%20%2Bc)
There is no bx in our given equation, so we put 0x
Given equation can be written as ![y=ax^2 + 0x +c](https://tex.z-dn.net/?f=y%3Dax%5E2%20%2B%200x%20%2Bc)
a=a , b=0
Now we use formula to find vertex
![x=\frac{-b}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%7D%7B2a%7D)
![x=\frac{-0}{2a}=0](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-0%7D%7B2a%7D%3D0)
Now we plug in 0 for 'a' and find out y
![y=a(0)^2 + 0x +c= c](https://tex.z-dn.net/?f=y%3Da%280%29%5E2%20%2B%200x%20%2Bc%3D%20c)
So our vertex is (0,c)
The axis of symmetry at x coordinate of vertex
So x=0 is our axis of symmetry
Well if the first point is (-2, 1) and the slope is three than the other point would be (-1, 4)