What is the vertex form of $y = x^{2} - 8 x + 16$ $y=x^2 - 8x + 16$ ?

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Answer 1 · 2016-02-18T02:29:26

It is $y = {(x - 4)}^{2}$ $y=(x-4)^2$

The vertex form of a parabola's equation is generally expressed as :

$y = a \cdot {(x - h)}^{2} + k$ $y=a*(x-h)^2+k$

Hence the given parabola can be written as follows

$y = {(x - 4)}^{2}$ $y=(x-4)^2$

so it is $a = 1$ $a=1$ , $h = 4$ $h=4$ , $k = 0$ $k=0$

So the vertex is $(h = 4, k = 0)$ $(h=4,k=0)$

graph{(x-4)^2 [-1.72, 12.33, -0.69, 6.333]}

What is the vertex form of y=x^2 - 8x + 16 y=x2−8x+16y=x^2 - 8x + 16 ?