Question #64cd4

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Question #64cd4

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Answer 1 · 2017-10-16T23:05:00

Vertex: $(- 3, - 6)$ $(-3, -6)$ , minimum
Domain: $(- \infty, \infty)$ $(-oo, oo)$
Range: $(- 6, \infty)$ $(-6, oo)$

Explanation:

This parabola is written in vertex form, where $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ . This form makes it simple to find what you need to know. The parabola has vertex $(h, k)$ $(h, k)$ , which in this problem is $(- 3, - 6)$ $(-3, -6)$ . Because $a$ $a$ is positive, in this case just $1$ $1$ , we know the parabola opens upward. Therefore, the vertex is the minimum point on the parabola. The domain of any parabola (when $y$ $y$ is a function of $x$ $x$ ) is $(- \infty, \infty)$ $(-oo, oo)$ or all real numbers. Since, the parabola has a minimum height at $y = - 6$ $y=-6$ and expands upward forever, its range is $(- 6, \infty)$ $(-6, oo)$ .

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Question #64cd4

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