Question

Sketch the graph of y = 5 sin 2x° + 12​

Answer 1

Step-by-step explanation:

$\displaystyle \boxed{y = 5cos\:(2x - \frac{\pi}{2}) + 12} \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 12 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{\pi}{4}} \hookrightarrow \frac{\frac{\pi}{2}}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 5$

OR

$\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 12 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 5$

You will need the above information to help you interpret the graph. Now, if you plan on writing your equation as a function of cosine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of $\displaystyle y = 5cos\:2x + 12,$ in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the sine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the cosine graph [photograph on the right] is shifted $\displaystyle \frac{\pi}{4}\:unit$to the left, which means that in order to match the sine graph [photograph on the left], we need to shift the graph FORWARD $\displaystyle \frac{\pi}{4}\:unit,$which means the C-term will be positive, and by perfourming your calculations, you will arrive at $\displaystyle \boxed{\frac{\pi}{4}} = \frac{\frac{\pi}{2}}{2}.$So, the cosine graph of the sine graph, accourding to the horisontal shift, is $\displaystyle y = 5cos\:(2x - \frac{\pi}{2}) + 12.$Now, with all that being said, in this case, sinse you ONLY have a wourd problem to wourk with, you MUST use the above formula for calculating the period. Onse you do that, the rest should be easy. Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at $\displaystyle y = 12,$in which each crest is extended five units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

Answer 2

Function: y = 5 sin (2(x))+ 12

Find y-intercept:

y = 5 sin 2(0)+ 12

y = 12

​$\sf Y\:Intercepts}:\:\left(0,\:12\right)$

→ Formula for maximum: M = A + |B|

Maximum:

12 + |5|

17

When y = 17

$\rightarrow \sf 17 = 5 sin (2(x))+ 12$

$\rightarrow \sf \sf 5 sin (2(x)) = 5$

$\rightarrow \sf sin (2(x)) = 1$

$\rightarrow \sf 2x = sin^{-1}(1)$

$\rightarrow \sf 2x = 90^{\circ \:}, \ \ 450^{\circ \:}$

$\rightarrow \sf x = 45^{\circ \:}, \ \ 225^{\circ \:}$

maximum: ( 45° , 17 ), (225° , 17), .....

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→  Formula for minimum: m = A ‐ |B|

Minimum:

12 - |5|

7

When y = 7

$\rightarrow \sf 7 = 5 sin (2(x))+ 12$

$\rightarrow \sf 5 sin (2(x)) = -5$

$\rightarrow \sf 2(x) = sin^{-1}(-1)$

$\rightarrow \sf x = -45^{\circ \:} , \ \ 135^{\circ \:}$

minimum: ( -45°,7), (135°, 7), .....

Repeat the same process for finding more values on the x-axis, or just follow the trend of the curve from the points found and sketch the graph easily.

$\sf Domain\:\left(-\infty \: < x < \infty )$

$\sf Range : 7\le \:f\left(x\right)\le \:17$

Sketched below: