Question

What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola `y = –3(x + 8)^2 + 5`?

Answer 1

1) `(-8,5)`
2) `x=-8`
3) max = `5`, min = `-infty`
4) R = `(-infty,5]`

Explanation:

1) let's traslate:

`y'=y`
`x'=x-8`

so the new parabola is `y'=-3x'^2+5`

the vertex of this parabola is in `(0,5) =>` the vertex of the old parabola is in `(-8,5)`

NB: you could have solve this even without the translation, but it would have been just a waste of time and energy :)

2) The axis of symmetry is the vertical lie passing through the vertex, so `x=-8`

3) It is a downward-facing parabola because the directive coefficient of the quadratic polynomial is negative, so the max is in the vertex, i.e. max=5, and the minimum is `-infty`

4) Because it is a continuous function, it satisfies Darboux property so the range is `(-infty,5]`

NB: If you don't know Darboux property, it is trivial to prove that if `exists y_0 < y_1 : exists x_0 and x_1 : y_0=-3(x_0+8)^2+5` and `y_1=-3(x_0+8)^2+5`, so `forall y in (y_0,y_1) exists x : y=-3(x+8)^2+5`, you just have to solve the equation and use the relations to prove that `Delta>=0`