What is the vertex of $y = 4 x - x^{2}$ $y=4x-x^2$ ?

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Answer 1 · 2018-06-23T09:34:19

(2,4)

$\frac{- b}{2 a}$ $(-b)/(2a)$ gives the X coordinate of the vertex

$\frac{- 4}{2 \times - 1} = \frac{- 4}{- 2} = 2$ $(-4)/(2xx-1)=(-4)/(-2)=2$

Put x=2 into $y = 4 x - x^{2}$ $y=4x-x^2$

$y = 4 \times 2 - 2^{2}$ $y=4xx2-2^2$

$y = 8 - 4 . \Rightarrow y = 4$ $y=8-4. => y =4$

What is the vertex of y=4x−x2y=4x-x^2?