How do you find the axis of symmetry, vertex and x intercepts for y=x^2-4x-1y=x2−4x−1?

1 Answer
Binayaka C.
Oct 24, 2017

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

Explanation:

y= x^2-4x-1 or y= (x^2-4x+4)-4-1y=x2−4x−1ory=(x2−4x+4)−4−1 or

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

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

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

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

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

0= (x-2)^2-5 or (x-2)^2=5 or (x-2)=+- sqrt5 0=(x−2)2−5or(x−2)2=5or(x−2)=±√5

:.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]

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