Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `y= x^2+ 5x - 12`?

Answer 1

`y = x^2+5x-12 = (x+5/2)^2-25/4-12`

`= (x+5/2)^2-73/4`

Hence vertex `(-5/2, -73/4)`, axis of symmetry `x=-5/2`, `y` intercept `(0, -12)`, `x` intercepts `(-5/2+-sqrt(73)/2, 0)`

Explanation:

`(x+5/2)^2 = x^2+5x+25/4`

So

`y = x^2+5x-12 = (x+5/2)^2-25/4-12`

`= (x+5/2)^2 - 73/4`

The minimum value of `(x+5/2)^2` occurs when `(x+5/2) = 0`, that is when `x = -5/2`

If `x+5/2 = 0` then `y = -73/4`.

So the vertex is at `(-5/2, -73/4)`

The parabola is vertical, so the axis of symmetry is just the vertical line through the vertex, i.e. `x=-5/2`.

The intercept with the `y` axis occurs where `x=0`. From the original equation we find `y = x^2-5x-12 = 0^2-5*0-12 = -12`. So the `y` intercept is at `(0, -12)`.

The intercepts with the `x` axis are where `y=0`. Then:

`(x+5/2)^2 - 73/4 = 0`

So `(x+5/2)^2 = 73/4`

So `x+5/2 = +-sqrt(73/4) = +-sqrt(73)/2`

and `x = -5/2+-sqrt(73)/2`

That is the `x` intercepts are `(-5/2+-sqrt(73)/2, 0)`