How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2+2x#?

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Answer 1 · 2017-07-20T12:25:25

Vertex is at # (-1, -1) # , y -intercept is #y=0 # and
axis of symmetry is #x = -1#

# f(x) = x^2 + 2x or f(x) = x^2 + 2x+1 -1 # or

#f(x) = (x+1)^2 -1 # , Comparing with standard vertex form of

equation #y= a(x-h)^2 +k ; (h,k)# being vertex , we find here

#h=-1, k=-1, a=1 # So vertex is at # (-1, -1) #

y -intercept can be found by putting #x=0# in the equation

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

Axis of symmetry is #x = -1#

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