Question

Use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions.

Answer 1

`y=-(x+1)^2+2`.

Explanation:

This is a parabola so it's parent is
`y=x^2`.

Let's described what happened to get from the parent to this.

The graph has been reflected so we will have
`y=-x^2`.

The graph has been moved left 1 and up 2 so this gives us:

`y=-(x-(-1))^2+2`.

Simplifying this gives us
`y=-(x+1)^2+2`.

Let's see if a few points we can identify can help confirm or convince you.

Some points I see that cross nicely are:

(-3,-2)

(-2,1)

(-1,2)

(0,1)

(1,-2)

Let's check them and see.

`y=-(x+1)^2+2` with
`(x,y)=(-3,-2)`:

`-2=-(-3+1)^2+2`

`-2=-(-2)^2+2`

`-2=-4+2`

`-2=-2` is true so (-3,-2) does satisfy
`y=-(x+1)^2+2`.

`y=-(x+1)^2+2` with
`(x,y)=(-2,1)`:

`1=-(-2+1)^2+2`

`1=-(-1)^2+2`

`1=-1+2`

`1=1` is true so (-2,1) does satisfy
`y=-(x+1)^2+2`.

`y=-(x+1)^2+2` with
`(x,y)=(-1,2)`:

`2=-(-1+1)^2+2`

`2=-(0)^2+2`

`2=0+2`

`2=2` is true so (-1,2) does satisfy
`y=-(x+1)^2+2`.

`y=-(x+1)^2+2` with
`(x,y)=(0,1)`:

`1=-(0+1)^2+2`

`1=-(1)^2+2`

`1=-1+2`

`1=-1` is true so (0,1) does satisfy
`y=-(x+1)^2+2`.

`y=-(x+1)^2+2` with
`(x,y)=(1,-2)`:

`-2=-(1+1)^2+2`

`-2=-(2)^2+2`

`-2=-4+2`

`-2=-2` is true so (1,-2) does satisfy
`y=-(x+1)^2+2`.

All the mentioned points satisfied our equation:

`y=-(x+1)^2+2`