Answer:
B(x+2)^2+y^2=16
Step-by-step explanation:
since you have an x-coordinate only and it is negative and a y coordinate which is 0..basically x will appear positive with its conjecture x^2 and the y value will be independent...y^2...bare in Mind that radius is always squared hence it is 16 instead of 4
Answer:
![\large\boxed{f(x)=2(x-1)^2+3}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7Bf%28x%29%3D2%28x-1%29%5E2%2B3%7D)
Step-by-step explanation:
The vertex form of a quadratic function f(x) = ax² + bx + c:
![f(x)=a(x-h)^2+k](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-h%29%5E2%2Bk)
Where (h, k) is a vertex.
![h=\dfrac{-b}{2a},\ k=f(h)](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B-b%7D%7B2a%7D%2C%5C%20k%3Df%28h%29)
We have:
![f(x)=2x^2-4x+5\to a=2,\ b=-4,\ c=5](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E2-4x%2B5%5Cto%20a%3D2%2C%5C%20b%3D-4%2C%5C%20c%3D5)
Calculate <em>h</em> and <em>k</em>:
![h=\dfrac{-(-4)}{2(2)}=\dfrac{4}{4}=1\\\\k=f(1)\to k=2(1^2)-4(1)+5=2-4+5=3](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B-%28-4%29%7D%7B2%282%29%7D%3D%5Cdfrac%7B4%7D%7B4%7D%3D1%5C%5C%5C%5Ck%3Df%281%29%5Cto%20k%3D2%281%5E2%29-4%281%29%2B5%3D2-4%2B5%3D3)
Finally:
![f(x)=2(x-1)^2+3](https://tex.z-dn.net/?f=f%28x%29%3D2%28x-1%29%5E2%2B3)
No picture is shown can’t tell
Answer:
B. 3a
Step-by-step explanation: