Question

How do you sketch the graph of `y=-2(x-1)^2+7` and describe the transformation?

Answer 1

See method below.

Explanation:

Transformation

Let `f(x)=x^2`

`f(x-1)=(x-1)^2`
This is a translation by 1 to the right.

`2f(x-1)=2(x-1)^2`
This is a vertical stretch, scale factor 2.

`-2f(x-1)=-2(x-1)^2`
This is a reflection in the x-axis (y=0).

`-2f(x-1)+7=-2(x-1)^2+7`
This is a translation up by 7.

Put this together; our transformation is:

Reflection in the x-axis, vetical stretch scale factor 2, transformation by `((1),(7))`.

Sketching

We need to find some points to plot the curve.

`y=-2(x-1)^2+7`

From this completed square form, we can tell that the minimum will be at `(1, 7)`

Find the x-intercepts:

Let `y=0`
`0=-2(x-1)^2+7`
`-7=-2(x-1)^2`
`7/2=(x-1)^2`
`+-sqrt(7/2)=x-1`
`x=1+-sqrt(7/2)`

so `x=1+sqrt(7/2)` or `x=1-sqrt(7/2)`

Find the y-intercepts:

`y=-2(x^2-2x+1)+7`
`y=-2x^2+4x-2+7`
`y=-2x^2+4x+5`

Let `x=0 => y=5`
So there is a y-intercept at `(0, 5)`

The co-efficient of `x^2` is negative, so it is an upside-down-U-shaped curve.

The graph should look like this:

graph{-2(x-1)^2+7 [-14.16, 17.88, -6.48, 9.54]}