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

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Answer 1 · 2018-06-15T10:56:34

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

#y = -x ^2 +4 x or y = -(x ^2 -4 x)# or

# y = -(x ^2 -4 x +4) +4# or

# y = -(x -2)^2 +4# Comparing with vertex form of

equation #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]