How do you find the vertex and the intercepts for $y = x^{2} - 2$ $y = x^2 - 2$ ?

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How do you find the vertex and the intercepts for $y = x^{2} - 2$ $y = x^2 - 2$ ?

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Answer 1 · 2017-11-03T12:00:06

vertex at $(x, y) = (0, - 2)$ $(x,y)=(0,-2)$
y-intercept: $y = 0$ $y=0$
x-intercepts: $x = - \sqrt{2}$ $x=-sqrt(2)$ and $x = + \sqrt{2}$ $x=+sqrt(2)$

Explanation:

One way to find the vertex is to convert the given equation into an explicit vertex form
$XXX y = {(x - a)}^{2} + b$ $color(white)("XXX")y=(x-color(red)a)^2+color(blue)(b)$ with vertex at $a, b)$ $color(red)a,color(blue)b)$

$y = x^{2} - 2$ $y=x^2-2$
is equivalent to
$XXX y = {(x - 0)}^{2} + (- 2)$ $color(white)("XXX")y=(x-color(red)0)^2+color(blue)((-2))$
the vertex form with vertex at $(0, (- 2))$ $(color(red)0,color(blue)((-2)))$

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

In this case $y = x^{2} - 2$ $y=color(green)x^2-2$ with $x = 0$ $color(green)x=color(green)0$
becomes $y = 0^{2} - 2 = - 2$ $y=color(green)0^2-2=-2$

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The x-intercepts are the values of $x$ $x$ which are possible when $y = 0$ $y=0$

In this case $y = x^{2} - 2$ $color(magenta)y=x^2-2$ with $y = 0$ $color(magenta)y=color(magenta)0$
becomes $0 = x^{2} - 2$ $color(magenta)0=x^2-2$
$XXX \to x^{2} = 2$ $color(white)("XXX")rarr x^2=2$
$XXX \to x = \pm \sqrt{2}$ $color(white)("XXX")rarr x=+-sqrt(2)$

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How do you find the vertex and the intercepts for $y = x^{2} - 2$ $y = x^2 - 2$ ?

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Explanation:

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How do you find the vertex and the intercepts for $y = x^{2} - 2$ $y = x^2 - 2$ ?