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

Jun 18, 2015

1) $\left(- 8 , 5\right)$
2) $x = - 8$
3) max = $5$, min = $- \infty$
4) R = $\left(- \infty , 5\right]$

#### Explanation:

1) let's traslate:

$y ' = y$
$x ' = x - 8$

so the new parabola is $y ' = - 3 x {'}^{2} + 5$

the vertex of this parabola is in $\left(0 , 5\right) \implies$ the vertex of the old parabola is in $\left(- 8 , 5\right)$

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 $\left(- \infty , 5\right]$

NB: If you don't know Darboux property, it is trivial to prove that if $\exists {y}_{0} < {y}_{1} : \exists {x}_{0} \mathmr{and} {x}_{1} : {y}_{0} = - 3 {\left({x}_{0} + 8\right)}^{2} + 5$ and ${y}_{1} = - 3 {\left({x}_{0} + 8\right)}^{2} + 5$, so $\forall y \in \left({y}_{0} , {y}_{1}\right) \exists x : y = - 3 {\left(x + 8\right)}^{2} + 5$, you just have to solve the equation and use the relations to prove that $\Delta \ge 0$