What is the vertex of the parabola $y = (x-4)^2$ ?

Question

1161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-22T15:47:14

$(4,0)$

Standard form; $" "y=ax^2+bx+c$
Vertex form; $" "y=a( x+ b/(2a))^2+k$

So your given equation is in vertex form in that we have:

$" "y=1(x-4)^2+0$

Where $x_("vertex")=(-1)xxb/(2a) -> (-1)xx(-4) = +4$

$" "y_("vertex")=k ->0$

$color(blue)("Vertex "->(x,y)->(4,0)$

What is the vertex of the parabola y = (x-4)^2y = (x-4)^2?