How do you find the vertex and intercepts for $y = 4 x - x^{2}$ $y=4x-x^2$ ?

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Answer 1 · 2018-06-15T10:56:34

Vertex is at $(2, 4)$ $(2,4)$ , x intercepts are at $(0, 0) and 4, 0)$ $(0,0) and 4,0)$ ,
y intercept is at $(0, 0)$ $(0,0)$

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

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

$y = - {(x - 2)}^{2} + 4$ $y = -(x -2)^2 +4$ Comparing with vertex form of

equation $f (x) = a {(x - h)}^{2} + k; (h, k)$ $f(x) = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=2 , k=4 :.$ Vertex is at $(2,4)$ y intercept

is found by putting $x=0$ in the equation $y = -x^2+4 x$ or

$y=0 :.$ y intercept is $y=0 or (0,0)$ x intercept is found by

putting $y=0$ in the equation $y = -x^2+4 x$ or

$0 = -x^2+4 x or x(4-x)=0:. x=0 and x=4$

x intercepts are at $(0,0) and 4,0)$

graph{-x^2+4 x [-10, 10, -5, 5]} [Ans]

How do you find the vertex and intercepts for y=4x-x^2y=4x−x2y=4x-x^2?