How do you find the vertex and the intercepts for #G(x) = x^2 + 4x#?

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Answer 1 · 2017-12-10T11:39:05

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

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

#G(x)=(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

#G(x)=x^2+4x # or #y=0 :.# y intercept is # (0,0)#

x intercepts are found by putting #y=0# in the equation

#G(x)=y=x^2+4x or x^2+4x =0 # or

#x(x+4) =0 :. x=0 and x+4=0 :. x =-4 #

x intercepts are at #(0,0) and (-4,0)#

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