How do you find the vertex of $f(x)=3x^2-25$ ?

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Answer 1 · 2016-10-07T19:55:47

$Vertex => (x,y) => (0,-25)$

Standard form of a quadratic equation

$ax^2+bx+c=0$

$x=-b/2a$

$y=f(-b/(2a))$

$Vertex => (x,y) => (-b/(2a),f(-b/(2a)))$

$f(x)=3x^2-25=3x^2+0x-25$

$a=3$
$b=0$
$c=-25$

Substitute

$x=-b/(2a)=-0/(2(3))=0$

$y=f(-b/(2a))=f(0)=3(0)^2-25=0-25=-25$

$Vertex => (x,y) => (0,-25)$

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How do you find the vertex of f(x)=3x^2-25f(x)=3x^2-25?