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

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Answer 1 · 2017-10-22T21:48:38

Vertex (h,k) = (2,4)

X-intercept (4,0), (0,0)

Y-intercept (0,0)

Vertex form of equation is
$y = a {(x - h)}^{2} + k$ $y = a (x-h)^2 + k$

$y = - x^{2} + 4 x = - (x^{2} - 4 x + 4) + 4 = - (1) {(x - 2)}^{2} + 4$ $y = -x^2 + 4x = - (x^2 -4x + 4) +4 = -(1)(x -2)^2 + 4$

$a = - 1, h = 2, k = 4$ $a = -1, h = 2, k = 4$

Vertex (h,k) = (2,4)

Substituting y = 0,
X-intercept (4,0), (0,0)

Substituting x = 0,
Y-intercept (0,0)

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