Question

Question #be5ca

Answer 1

The first one should look like this: graph{-2(x+2)^2+6 [-10, 10, -5, 5]}
The second one should look like this: graph{1/2(x)^2-6x+13 [-10, 10, -5, 5]}

Explanation:

Our method for graph the parabola will be like this:
We find the values of a, h, and k.
Then, we graph the other four points by doing the following:
We will do h-2a, h-a, h+a, and h+2a for the x values and plug them in the equation to find the y values.
The minimum/maximum point of a parabola is (h,k)

`-2(x+2)^2+6`:

This equation is in the form `a(x-h)^2+k`

Therefore,
`a=-2`
`h=-2`
`k=6`

The x values are:
`-2-2(-2)=2`
`-2-(-2)=0`
`-2`
`-2+(-2)=-4`
`-2+2(-2)=-6`

Plug in the values in the equation to get the following y values.
`-2-2(-2)=2` =>-28
`-2-(-2)=0` =>-2
`-2` =>6
`-2+(-2)=-4` =>-2
`-2+2(-2)=-6` =>-28

Plot the points an you have the answer!

`1/2(x)^2-6x+13`:

This equation is in the form `ax^2+bx+c`
We have to know that `h=(-b)/(2a)`
Plug in this value to the equation to get k.
Therefore,
`a=1/2`
`h=6`
`k=-5`

The x values are:
`6-2(1/2)=5`
`6-1/2=5 1/2`
`6`
`6+1/2=-6 1/2`
`6+2(1/2)=7`

Plug in the values in the equation to get the following y values.
`6-2(1/2)=5` =>-4.5
`6-1/2=5 1/2` =>-4.875
`6` =>-5
`6+1/2=-6 1/2`=>-4.875
`6+2(1/2)=7`=>-4.5

Plot the points an you have the answer!