How do you find the axis of symmetry and vertex point of the function: #y = x^2 − 1#?

Question

Eli Assouline · Accepted Answer

Rewrite in explicit vertex form and recognize it as a parabola in standard position.
Vertex:# (0,-1)color(white)("XXXX")#Axis of symmetry: #x=0# A parabola in standard position has the following vertex form: #color(white)("XXXX")y=m(x-color(red)(a))^2+color(blue)(b)# #color(white)("XXXXXXXX")#, with a vertex at #(a,b)# and an upward opening if #m>0#. The explicit vertex form of #y=x^2-1# is #color(white)("XXXX")y=1(x-color(red)(0))^2+color(blue)((-1))# #color(white)("XXXXXXXX")#, with a vertex at #(color(red)(0),color(blue)(-1))#. The axis of symmetry in standard position, #x=color(red)(0)#, is a vertical line that passes through the vertex.

Genesis Allen · Answer

undefined To find the axis of symmetry of a quadratic function in the form $ y = ax^2 + bx + c $, you can use the formula $ x = -\frac{b}{2a} $. For the given function $ y = x^2 - 1 $, $ a = 1 $ and $ b = 0 $. Plugging these values into the formula, we get $ x = -\frac{0}{2(1)} = 0 $. So, the axis of symmetry is $ x = 0 $.

To find the vertex point, substitute the value of $ x $ found for the axis of symmetry into the original function. So, when $ x = 0 $, $ y = (0)^2 - 1 = -1 $. Therefore, the vertex point is $ (0, -1) $.