How do I find the vertex of the function $- 6 {(x - 3)}^{2} - 5$ $-6(x-3)^2 -5$ ?

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Answer 1 · 2016-10-04T20:19:36

vertex $\to (x, y) = (3, - 5)$ $->(x,y)=(3,-5)$

This is what they call the vertex form of a quadratic.

Set as $y = - 6 {(x - 3)}^{2} - 5$ $" "y=-6(x color(red)(-3))^2color(green)(-5)$

From this x-vertex $= (- 1) \times - 3 = + 3$ $=(-1)xx color(red)(-3) = +3$
From this y-vertex $= - 5$ $=color(green)(-5)$

So vertex $\to (x, y) = (3, - 5)$ $->(x,y)=(3,-5)$

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How do I find the vertex of the function -6(x-3)^2 -5−6(x−3)2−5-6(x-3)^2 -5?