How do you find the vertex and the intercepts for $y = - x^{2} + 9$ $y= -x^2 + 9$ ?

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How do you find the vertex and the intercepts for $y = - x^{2} + 9$ $y= -x^2 + 9$ ?

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Answer 1 · 2017-02-09T12:18:15

Vertex: $(0, 9)$ $(0,9)$
Y-intercept: $y = 9$ $y=9$
X-intercepts: $x = - 3 and x = + 3$ $x=-3 and x=+3$

Explanation:

Remember the standard vertex form for a parabola is
$XXX y = m {(x - a)}^{2} + b$ $color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b$
with vertex at $(a, b)$ $(color(red)a,color(blue)b)$

Re-writing the given equation: $y = - x^{2} + 9$ $y=-x^2+9$
into explicit standard vertex form:
$XXX y = - 1 {(x - 0)}^{2} + 9$ $color(white)("XXX")y=color(green)(-1)(x-color(red)0)^2+color(blue)9$
with vertex at $(0, 9)$ $(color(red)0,color(blue)9)$

The Y-intercept is the value of $y$ $y$ when $x = 0$ $x=color(magenta)0$
$XXX y = - 0^{2} + 9 = 9$ $color(white)("XXX")y=-color(magenta)0^2+9=9$

The X-intercepts are the values of $x$ $x$ possible when $y = 0$ $y=color(orange)0$
$XXX 0 = - x^{2} + 9$ $color(white)("XXX")color(orange)0=-x^2+9$
$XXX \to x^{2} = 9$ $color(white)("XXX")rarr x^2=9$
$XXX \to x = \pm 3$ $color(white)("XXX")rarr x=+-3$

Here is the graph for verification purposes:
enter image source here

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How do you find the vertex and the intercepts for $y = - x^{2} + 9$ $y= -x^2 + 9$ ?

1 Answer

Explanation:

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How do you find the vertex and the intercepts for y= -x^2 + 9y=−x2+9y= -x^2 + 9?

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Explanation:

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How do you find the vertex and the intercepts for $y = - x^{2} + 9$ $y= -x^2 + 9$ ?