Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `y=-3x^2-12x-3`?

Answer 1

`y = f(x) = -3x^2-12x-3 = -3(x^2+4x+1)`

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

This has its vertex when `x+2 = 0`, that is when `x=-2`.

`f(-2) = -3((-2)^2+4(-2)+1) = -3(4-8+1) = -3xx-3 = 9`

So the vertex is at `(-2, 9)` and the axis of symmetry is the line `x = -2`

The `x`-axis intercepts are where `f(x) = 0`, which will happen when `(x+2)^2-3 = 0`, that is when

`(x+2)^2 = 3`

So

`x + 2 = +-sqrt(3)`

Subtract `2` from both sides to get

`x = -2 +- sqrt(3)`

The `y`-axis intercept is at `(0, f(0)) = (0, -3)`

graph{-3x^2-12x-3 [-20, 20, -10, 10]}