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

1 Answer
Jul 17, 2018

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)

Explanation:

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]