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