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2 years ago

Find the vertex I’d this parabola y=-2x^2-4x+4

Mathematics

1 answer:

mamaluj [8]2 years ago

5 0

Answer:

(-1, 6)

Step-by-step explanation:

A graphing calculator shows the vertex to be (x, y) = (-1, 6).

You can also find it by putting the equation into vertex form.

y = -2(x^2 +2x) +4

y = -2(x^2 +2x +1) +4 +2(1) . . . . . add 1 inside parens; the opposite outside

y = -2(x +1)^2 +6

Compare to ...

y = a(x -h)^2 +k

and you see a=-2, h=-1, k=6

The vertex is (h, k) = (-1, 6).

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Step-by-step:

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serg [7]

Answer:

The result of the integral is:

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Step-by-step explanation:

We are given the following integral:

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Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

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$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$