What is the the vertex of $x = (y -3)^2 - 5y-5$ ?

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Answer 1 · 2018-07-05T20:35:36

$11/2,-105/4$

Let $f(y)=(y-3)^2-5y-5$
then we get by using

$(a-b)^2=a^2-2ab+b^2$

$f(y)=y^2-6y+9-5y-5$
combining like terms

$f(y)=y^2-11y+4$

we compute the coordinates of the vertex:_

$f'(y)=2y-11$
so

$f'(y)=0$ if $y=11/2$

and
$f(11/2)=-105/4$

What is the the vertex of x = (y -3)^2 - 5y-5 x = (y -3)^2 - 5y-5 ?