Question

How do you find the axis of symmetry, vertex and x intercepts for `y=x^2-4x`?

Answer 1

Axis of symmetry = 2
Vertex = (2,-4)
X-intercepts = 0,4

Explanation:

The axis a symmetry can be found using a simple formula of

`x=-b/(2a)`

when the graph is in the form of

`y=ax^2+bx+c`

In your graph, `a` would equal 1 and `b` would equal -4. Substituting this into the equation would give you the axis of symmetry of

`x=(-(-4))/(2(1)`

`x=2`

Additionally, the vertex also resides on the axis of symmetry. We just need to find the y-coordinate of the vertex, which we can do by substituting `x=2` back into the orginal equation giving us

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

`y=-4`

Thus, the vertex will lie on `(2,-4)`

To find the x-intercepts, we must find where the graph intersects the x-axis and this can be done when we let the graph equal zero.

`0=x^2-4x`

Factorising we get

`0=x(x-4)`

Finally, using the null factor law, we can find what the x-intercepts are.

`x=0,4`

Answer 2

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

Explanation:

`y = x^2 -4x = x^2 -4x +4 - 4 = (x-2)^2 -4` Comparing with

Standard vertex fom of equation `y = a (x-h)^2 ; (h,k)` being

vertex we find here `h = 2 , k= -4`, so vertex is at `(2,-4)`.

Axis of symmetry is `x = 2 ; x` intercepts can be found by putting

`y=0` in the equation .

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

`:. x=0 or x=4` So `x` intercepts are `x=0 ,x=4`

Vertex is at `(2,-4)` , axis of symmetry is `x = 2` and

`x` intercepts are `x=0 ,x=4`.

graph{x^2-4x [-10, 10, -5, 5]} [Ans]