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dezoksy [38]
3 years ago
6

Given the function g(x)=x^2-2 find the range for the domain (-2,-1,1,3)

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

(-2,2)

(-1,-1)

(1,-1)

(3,7)

If you graph it with a graphing calc. and pinpoint the X's(domains) then you can see the Y's(ranges)

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Rewrite the equation in vertex form. then find the vertex of the graph. y=-3x^2-5x+1
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Answer: y = -3(x + \frac{5}{6})² + \frac{37}{12}, (-\frac{5}{6}, \frac{37}{12})

<u>Step-by-step explanation:</u>

First, you need to complete the square:

y   = -3x² - 5x + 1

<u> -1  </u>   <u>                -1  </u>

y - 1 = -3x² - 5x

y - 1 = -3(x² + \frac{5}{3}x

y - 1 + -3(\frac{25}{36}) = -3(x² + \frac{5}{3}x + \frac{25}{36})

           ↑                     ↓            ↑

                                  \frac{5}{3*2} = (\frac{5}{3*2})^{2}

y - 1 - \frac{25}{12} = -3(x + \frac{5}{6})²

y - \frac{12}{12} - \frac{25}{12} = -3(x + \frac{5}{6})²

y  - \frac{37}{12} = -3(x + \frac{5}{6})²

y = -3(x + \frac{5}{6})² + \frac{37}{12}

Now, it is in the form of y = a(x - h)² + k   <em>where (h, k) is the vertex</em>

Vertex = (-\frac{5}{6}, \frac{37}{12})

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