Question

What is the axis of symmetry and vertex for the graph `y=x^2-3x+8`?

Answer 1

Vertex `(3/2, 23/4)`
Axis of symmetry: `x=3/2`

Explanation:

Given a quadratic of the form `y=ax^2+bx+c` the vertex, `(h,k)` is of the form `h=-b/(2a)` and `k` is found by substituting `h`.

`y=x^2-3x+8` gives `h=-(-3)/(2*1)=3/2`.

To find `k` we substitute this value back in:

`k=(3/2)^2-3(3/2)+8 = 9/4-9/2+8 = 23/4`.

So the vertex is `(3/2, 23/4)`.

The axis of symmetry is the vertical line through the vertex, so in this case it is `x=3/2`.