Question

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

Answer 1

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

`=(x-1/2)^2-17/4`

The vertex is at `(1/2, -17/4)`

The axis of symmetry is the vertical line `x = 1/2`

The intercept with the `y` axis is `(0, -4)`

The intercepts with the `x` axis are the points where `y=0`, so

`(x-1/2)^2 = 17/4`

`(x-1/2) = +- sqrt(17/4) = +-sqrt(17)/2`

So

`x = 1/2+-sqrt(17)/2 = (1+-sqrt(17))/2`

That is `((1-sqrt(17))/2, 0)` and `((1+sqrt(17))/2, 0)`

Answer 2

x of vertex: `x = (-b/2a) = 1/2`

y of vertex: `f(1/2) = 1/4 - 1/2 - 4 = -17/4`

Axis of symmetry: x = (-b/2a) = 1/2

x-intercepts: y = x^2 - x - 4 = 0

`D = d^2 = 1 + 16 = 19 -> d = +- sqrt17`

`x = 1/2 +- sqrt17/2`