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2 years ago

Write a formula for quadratic function if its graph has the vertex at point (0,6) and passes through the point (−1,−2).

Mathematics

1 answer:

d1i1m1o1n [39]2 years ago

4 0

Answer:

y = - 8x² + 6

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (0, 6), thus

y = a(x - 0)² + 6, that is

y = ax² + 6

To find a substitute (- 1, - 2) into the equation

- 2 = a(- 1)² + 6, that is

- 2 = a + 6 ( subtract 6 from both sides )

- 8 = a

y = - 8x² + 6

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Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

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Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

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3 years ago

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2 years ago

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irina [24]

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krek1111 [17]

Answer:

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Step-by-step explanation:

to solve this, we need to get rid of the brackets as follow:

3a2+1-4-2a2

ading the common terms we have:

a2 - 3

that is all.

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2 years ago