Sketch the graph. Label the vertex and axis of symmetry and is it wide narrow or normal? y = -x^2. I'm not sure how to do this... Help!?

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embizze's profile pic

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

The equation for a general quadratic is `y=ax^2+bx+c` . The shape of the graph is a parabola.

The vertex has x-coordinate `x=-b/(2a)` and the equation for the axis of symmetry is also `x=-b/(2a)` .

For `y=-x^2` we have a=-1,b=c=0. The axis of symmetry is x=0. The vertex has x-coordinate 0, so y=0 and the vertex is at (0,0).

The graph is the same as that of `y=x^2` only reflected across the x-axis.

The graph:

The vertex is at (0,0) and the axis of symmetry is the y-axis or x=0.

atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

The formula you use to find the axis is `-b / 2a`    a=-1 b=0 c=0       `0 / 2(-1) `   =   `0 / -2`   =    0

x=0   to find the y plug it into the equation `y= -0^2`  =0  so the vertex    is  `(0,0)`  and the axis of symmetry is always the x value of the vertex `aos=0`  

it is wide because when a is less than one it opens wider. 

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