Question

How do you graph `f(x) = 3 - abs(2x + 3)`?

Answer 1

This is an inverted V shape, with sides of slope `+-2`, vertex at `(-3/2, 3)`, intersection with the `y` axis at `(0, 0)` and intersections with the `x` axis at `(0, 0)` and `(-3, 0)`.

graph{3-abs(2x+3) [-10, 10, -5, 5]}

Explanation:

When `x <= -3/2`, `2x+3 <= 0` so `abs(2x+3) = -2x-3`

and `f(x) = 3-abs(2x+3) = 3 - (-2x-3) = 2x+6`

So this part of the curve is a line of slope `2` which intersects the `x` axis where `2x+6 = 0`, that is at `(-3, 0)`

When `2x+3 = 0`, `x = -3/2` and `f(x) = 2x+6 = 3`.

So the point `(-3/2, 3)` is the minimum of `abs(2x+3)` and the maximum of `f(x)`, thus the vertex.

When `x >= -3/2`, `2x+3 >= 0` so `abs(2x+3) = 2x+3`

and `f(x) = 3-abs(2x+3) = -2x`

So this part of the curve is a line of slope `-2` which intersects the `x` and `y` axis at `(0, 0)`.