What is the the vertex of $x = {(y + 2)}^{2} - 5$ $x = (y +2)^2 - 5$ ?

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Answer 1 · 2016-04-07T11:11:04

$vertex \to (x, y) \to (- 5, - 2)$ $color(blue)("vertex "-> (x,y) ->(-5,-2)$

This is a transformed quadratic: rotated clockwise by $\frac{π}{2} \to 90^{o}$ $pi/2->90^o$

So you swap round the x's and y's

$If it was a standard quadratic then vertex \to (x, y) \to (- 2, - 5)$ $color(green)("If it was a standard quadratic then vertex"-> (x,y)-> (-2,-5))$

$But we have to swap the values round so we have:$ $color(brown)("But we have to swap the values round so we have:")$

$vertex \to (x, y) \to (- 5, - 2)$ $color(blue)("vertex "-> (x,y) ->(-5,-2)$
Tony B

What is the the vertex of x=(y+2)2−5x = (y +2)^2 - 5 ?