How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y=x^2 +2x -5#?

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Answer 1 · 2018-07-17T11:49:18

Vertex is at # (-1,-6.)# , axis of symmetry is #x=-1#
y-intercept is at #(0,-5)#, x-intercepts are at #(-3.45,0) and (1.45,0)#

#y=x^2+2 x-5 # or

#y=(x^2+2 x +1)-1-5 # or

#y=(x+1)^2-6 # . This is vertex form of equation

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

#:.h=-1 ,k=-6,a=1 ; a>0 :.#Parabola opens upward.

Therefore vertex is at # (-1,-6.)# . Axis of symmetry is

#x= h or x = -1 # . Graphing: y-intercept is at

#x=0:.y= x^2+2 x-5= -5 or (0,-5) #, x-intercepts are

at #y=0 ; y=(x+1)^2-6 or (x+1)^2-6=0 # or

#(x+1)^2-6=0 or (x+1)^2=6 or (x+1)= +-sqrt6 # or

#x= -1 +-sqrt 6 or x~~1.45 , x = -3.45# Therefore,

x-intercepts are at #(-3.45,0) and (1.45,0)#

graph{x^2+2x-5 [-20, 20, -10, 10]} [Ans]