Is this written correctly?
Answer: you just need to do A=bh
Step-by-step explanation:
So you need to multiply the base by the height
Answer:
Vertex form: ![f(x)=(x-1)^2-3](https://tex.z-dn.net/?f=f%28x%29%3D%28x-1%29%5E2-3)
Standard form: ![f(x)=x^2-2x-2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2-2x-2)
Step-by-step explanation:
A quadratic function in vertex form is
where
is the vertex.
We are given
.
Let's plug that in:
.
Now let's find
.
We will use the
-intercept
to find
.
![f(0)=a(0-1)^2-3](https://tex.z-dn.net/?f=f%280%29%3Da%280-1%29%5E2-3)
![-2=a(0-1)^2-3](https://tex.z-dn.net/?f=-2%3Da%280-1%29%5E2-3)
![-2=a(-1)^2-3](https://tex.z-dn.net/?f=-2%3Da%28-1%29%5E2-3)
![-2=a(1)-3](https://tex.z-dn.net/?f=-2%3Da%281%29-3)
![-2=a-3](https://tex.z-dn.net/?f=-2%3Da-3)
![1=a](https://tex.z-dn.net/?f=1%3Da)
So the function in vertex form is:
.
In standard form, we will have to multiply and combine any like terms.
Let's do that:
![f(x)=(x-1)(x-1)-3](https://tex.z-dn.net/?f=f%28x%29%3D%28x-1%29%28x-1%29-3)
![f(x)=x^2-x-x+1-3](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2-x-x%2B1-3)
![f(x)=x^2-2x-2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2-2x-2)