Question

How do you find the axis of symmetry and vertex point of the function: `y=x^2-6x+4`?

Answer 1

Axis of symmetry: `x=3`
Vertex is at `(3,-5)`.

Explanation:

Notice that
`x^2−6x+4 = x^2−6x+9 - 5 = (x-3)^2 - 5`

Therefore, the graph of function
`y = x^2−6x+4`
can be obtained from the graph of function `y = x^2` by shifting it by `3` units to the right (that would be a graph of `y = (x-3)^2`) and then shifting it by `5` units down (to get to a graph of `y = (x-3)^2−5`).

graph{x^2-6x+4 [-7, 10, -7, 7]}

The above transformation shifts the axis of symmetry of function `y=x^2` from the position coinciding with Y-axis to the right by `3` units and it shifts the vertex of function `y=x^2`, firstly, to the right by `3` and then down by `5`.

Now we know that the axis of symmetry is a vertical line parallel to Y-axis that intersects X-axis at point `x=3`.

We also know that the vertex of the parabola is at point `(3,-5)`