Question

How do you find the intercepts, vertex and graph `f(x)=x^2-4x+4`?

Answer 1

We have only one `x`-intercept at `(2,0)` and `y`-intercept is at `(0,4)`. Vertex is at `(2,0)`.

Explanation:

For finding intercepts and vertex for an equation of the form `y=ax^2+bx+c`, we should convert it to form `y=a(x- alpha)(x- beta)`, where `x`-intercepts are `alpha` and `beta`, while `y`-intercept is `aalphabeta`, obtained by putting `x=0`; and to find intercept and to vertex form `y=a(x- h)^2+k` to find vertex `(h,k)`.

As `y=x^2-4x+4`

= `(x-2)^2`

Observe that this equation `y=(x-2)^2` is both in intercept form as well as vertex form. As `y=(x-2)^2`, intercepts is just `2` at `(2,0)`. `y`-intercept is where `x=0` i.e. `y=4` and at `(0,4)`.

Further `y=(x-2)^2` can also be written as

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

Vertex is `(2,0)`.

graph{ (x-2)^2 [-5, 5, -0.5, 4.5]}