What is the axis of symmetry and vertex for the graph $y = - x^{2} + 1$ $y= -x^2 + 1$ ?

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Answer 1 · 2017-04-08T14:49:01

Axis of symmetry is $x = 0$ $x=0$ ( $y$ $y$ -axis) and vertex is $(0, 1)$ $(0,1)$

The axis of symmetry of $(y - k) = a {(x - h)}^{2}$ $(y-k)=a(x-h)^2$ is $x - h = 0$ $x-h=0$ and vertex is $(h, k)$ $(h,k)$ .

As $y = - x^{2} + 1$ $y=-x^2+1$ can be written as

$(y - 1) = - 1 {(x - 0)}^{2}$ $(y-1)=-1(x-0)^2$

hence axis of symmetry is $x - 0 = 0$ $x-0=0$ i.e. $x = 0$ $x=0$ ( $y$ $y$ -axis) and vertex is $(0, 1)$ $(0,1)$

graph{-x^2+1 [-10.29, 9.71, -6.44, 3.56]}

Note: The axis of symmetry of $(x - h) = a {(y - k)}^{2}$ $(x-h)=a(y-k)^2$ is $y - k = 0$ $y-k=0$ and vertex is $(h, k)$ $(h,k)$ .

What is the axis of symmetry and vertex for the graph y=−x2+1y= -x^2 + 1?