Question

How to graph a parabola `y=3/2x^2 - 3x-5/2`?

Answer 1

Complete the square to find the vertex, axis of symmetry, etc.

Explanation:

`3/2(x-1)^2 = 3/2(x^2-2x+1) = 3/2x^2-3x+3/2`

So

`y = 3/2x^2-3x-5/2`

`= 3/2(x-1)^2-3/2-5/2`

`= 3/2(x-1)^2-4`

This is in vertex form,

allowing us to tell that the vertex is at `(1, -4)`

and the axis of symmetry is `x=1`.

The intersection with the `y` axis is at `(0, 3/2)`

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

`0 = 3/2(x-1)^2 - 4`

Add `4` to both sides to get:

`3/2(x-1)^2 = 4`

Multiply both sides by `2/3` to get:

`(x-1)^2 = 2/3*4`

Hence `x = 1+-2sqrt(2/3) ~= 1+-1.633`

So the intersections with the `y` axis are approximately:

`(-0.633,0)` and `(2.633, 0)`

graph{3/2x^2-3x-5/2 [-10, 10, -5, 5]}