Question

How do you find its vertex, axis of symmetry, y-intercept and x-intercept for `f(x) = x^2 - 4x`?

Answer 1

Vertex is `(2,-4)`; axis of symmetry is `x-2=0`; `x`-intercepts are `0` and `4` and `y`-intercept is `0`.

Explanation:

This is the equation of parabola. Intercept form of equation of parabola is `f(x)=a(x-alpha)(x-beta)`, where `alpha` and `beta` are intercepts on `x`-axis.

As we have `f(x)=x^2-4x=x(x-4)=(x-0)(x-4)`, `x`-intercepts are `0` and `4`.

`y`-intercept is obtained by putting `x=0` and hence it is `0^2-4*0=0-0=0` and hence `y`-intercept is `0`.

Vertex form of equation of parabola is `f(x)=a(x-h)^2+k`, where `(h,k)` is vertex and `x-h=0` is axis of symmetry

Here we have `f(x)=x^2-4x=(x^2-4x+4)-4=(x-2)^2-4`

Hence axis of symmetry is `x-2=0` and vertex is `(2,-4)`.

graph{(x^2-4x-y)(x-2)((x-2)^2+(y+4)^2-0.03)=0 [-9.92, 10.08, -5.12, 4.88]}