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Mariulka [41]
3 years ago
11

Find the y-intercept for the parabola defined by the equation y=x^2-x-12

Mathematics
1 answer:
o-na [289]3 years ago
6 0

Answer:

The y-intercept is (0,-12).

Step-by-step explanation:

Given equation is y=x^2-x-12.

Now question says to find the y-intercept of the given parabola y=x^2-x-12.

We know that y-intercept means the y-value when x-value equals 0

so let's plug x=0 into given equation y=x^2-x-12  then solve that for y.

y=x^2-x-12

y=0^2-0-12

y=0-0-12

y=-12

Hence the y-intercept is (0,-12).

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titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

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\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

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4

π

]∪[

4

3π

;π]

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4

π

]

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4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

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Display

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