How do you find the axis of symmetry, vertex and x intercepts for #y=x^2-4x-1#?

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Answer 1 · 2017-10-24T11:50:37

Axis of symmetry is # x =2 #, vertex is at # (2, -5)# and
x-intercepts are at #( -0.236,0) and (4.236,0)#

# y= x^2-4x-1 or y= (x^2-4x+4)-4-1# or

#y= (x-2)^2-5# . This is vertex form of

equation #y=a(x-h)^2+k ; a=1 ,h=2 ,k= -5 #

Therefore vertex is at #(h,k) or (2, -5)#

Axis of symmetry is #x= h or x =2 # , x-intercepts

is found by putting #y=0# in the equation #y= (x-2)^2-5# or

#0= (x-2)^2-5 or (x-2)^2=5 or (x-2)=+- sqrt5 #

#:.x=2 +sqrt5 and x=2-sqrt5 # or

#x~~4.236 (3dp) and x ~~ -0.236(3dp) # Hence

x-intercepts are at #( -0.236,0) and (4.236,0)#

graph{x^2-4x-1 [-10, 10, -5, 5]} [Ans]