What is the the vertex of $y = {(x - 16)}^{2} + 40$ $y =(x - 16)^2 + 40$ ?

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What is the the vertex of $y = {(x - 16)}^{2} + 40$ $y =(x - 16)^2 + 40$ ?

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Answer 1 · 2015-12-23T09:38:41

$(16, 40)$ $(16,40)$

Explanation:

The equation is already in vertex form so it is possible to see that the value is at its minimum when the squared factor is zero. this occurs when $x = 16$ $x = 16$ and at that point $y = 40$ $y= 40$

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What is the the vertex of $y = {(x - 16)}^{2} + 40$ $y =(x - 16)^2 + 40$ ?

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Explanation:

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Humanities

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What is the the vertex of y =(x - 16)^2 + 40 y=(x−16)2+40y =(x - 16)^2 + 40 ?

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Explanation:

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What is the the vertex of $y = {(x - 16)}^{2} + 40$ $y =(x - 16)^2 + 40$ ?