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How do find the vertex and axis of symmetry, and intercepts for a quadratic equation $y = - 2 x^{2}$ $y= -2x^2$ ?

1 Answer

Aug 21, 2015

vertex at $(0, 0)$ $(0,0)$
axis of symmetry: $x = 0$ $x=0$
y-intercept: $0$ $0$
x-intercept: $0$ $0$

Explanation:

General vertex form for a quadratic is
$XXXX y = m (x - a) + b$ $color(white)("XXXX")y=m(x-a)+b$
$XXXXXXXX$ $color(white)("XXXXXXXX")$ with the vertex at $(a, b)$ $(a,b)$

$y = - 2 x^{2}$ $y=-2x^2$ can be written in vertex form as
$XXXX y = (- 2) {(x - 0)}^{2} + 0$ $color(white)("XXXX")y = (-2)(x-0)^2+0$
$XXXXXXX$ $color(white)("XXXXXXX")$ with vertex at $(0, 0)$ $(0,0)$

The equation is that of a standard parabola (opening downward)
so the axis of symmetry is a vertical line passing through the vertex
i.e. $x = 0$ $x=0$

The y-intercept is the value of $y$ $y$ when $x = 0$ $x=0$

The x-intercept(s) is/are the value of $x$ $x$ when $y = 0$ $y=0$ (in this case there is only one solution to $0 = - 2 x^{2}$ $0=-2x^2$ )

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