Question

How do you use the important points to sketch the graph of `y=(x-2)^2`?

Answer 1

See below

Explanation:

`y=(x-2)^2`

`= x^2-4x+4`

Notice that `y` is a quadratic function of the form `ax^2+bx+c`, therefore its graph will be a parabola.

`y =0` at `x=2` - Notice that this is a coincident root meaning that `y` will touch the `x-`axis at `x=2` and at nowhere else.

Since `a =+1>0, y` will have a minimum value at
`x=(-b)/(2a) = 4/2 =2`

`:. y_min = 0` at `x=2`

Hence an "important point" is `(2,0)` being both the only `x-`intercept and the minimum of `y`

Next, we can find the `y-`intercept, where `x=0`

`y(0) = (-2)^2 =4`

So another "important point" is `(0,4)` the `x-`intercept

We can see these points on the graph of `y` below.

graph{(x-2)^2 [-6.83, 10.945, -1.68, 7.21]}