How do you find the axis of symmetry, vertex and x intercepts for `y=x^2-5x+6`?

First, to find the axis of symmetry, we need to find whether this function is even or odd.

A function is even if `f(x)=f(-x)`
A function is odd if `-f(x)=f(-x)`

If a function is even, it is symmetric to the y axis. If it is odd, it is symmetric to the origin. Also, it is possible for a function to be neither.

Now, we test the possibilities.
If `f(x)=x^2-5x+6`, then `f(-x)=x^2+5x+6` and `-f(x)=-x^2+5x-6`

We see that none of them are equal to each other. Therefore, this function is not symmetric to any axis.

To find the vertex of a quadratic function, we need to find the `(h,k)`. To do this, remember that `h=(-b)/(2a)` and `k` can be found if you plug `h` into the function. This is a useful trick if the function is in the form of `ax^2+bx+c=y`

Therefore, `h=5/2`

Plugging this into our equation, we get `k=-1/4`

The vertex is `(5/2,-1/4)`

To find the x intercepts, we need to find the zeros of this function.Zeros of the function are the values of `x` that when put into the function, results `y=0`.

Now, we need to solve the equation `x^2-5x+6=0`

We can factor this to get `(x-2)(x-3)=0`

This means that the x intercepts are `(2,0)` and `(3,0)`