Answer:
[tex] - \frac{8}{\pi} [/tex]
Step-by-step explanation:
Use slope formula.
Rise over run.
Or
y over x.
[tex] \frac{ {y}^{2} - y {}^{1} }{ {x}^{2} - x {}^{1} } [/tex]
Over changes in x value would be - 3/4 pi.
Plug in seepage intervals for x to find y.
[tex]6 \cos(2(\pi) - 4[/tex]
In the regular function,
[tex] \cos(\pi) = 1[/tex]
Since our period is 2, it would stay the same since 1x2=2
Since our amplitude is 6, our y value now is 6.
Since our vertical shift is -4, our y value is 2.
So
[tex]6 \cos(2(\pi)) - 4 = 2[/tex]
[tex] {x}^{2} = \pi \: \: \: \: {y}^{2} = 2[/tex]
Let do the other point,
[tex] \cos( \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2} [/tex]
Our period is 2 so
[tex] \frac{ {\pi} }{4} \times \frac{2}{1} = \frac{\pi}{2} [/tex]
[tex] \cos( \frac{\pi}{2} ) = 0[/tex]
Multiply this by 6.
It stays 0 then subtract 4 we get
[tex]6 \cos(2( \frac{\pi}{4} )) - 4 = - 4[/tex]
Use the earlier formula, slope
[tex] \frac{2 + 4}{ - \frac{3\pi}{4} } = - \frac{8}{\pi} [/tex]
