joshwelch17
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Chris wanted to transform the graph of the parent function Y= cot (x) by horizontally compressing it so that it has a period of 2/π units, horizontally Terslating it π/4 units to the right, and vertically translating it 1 unit up. To do so, he graphed the function y= cot (2x-π/4)+1 as shown. What did he do wrong?​

Respuesta :

kudzordzifrancis

Answer:

He didn't calculate the b-value correctly.

Step-by-step explanation:

The given parent function is:

[tex]y = \cot(x) [/tex]

The transformation is of the form:

[tex]y =a \cot(bx + c) + d[/tex]

The period is given by

[tex] \frac{\pi}{b} [/tex]

If we want the new function to have a period of

[tex] \frac{2}{\pi} [/tex]

Then we solve the following equation for b.

[tex] \frac{\pi}{b} = \frac{2}{\pi} [/tex]

[tex]b = \frac{ {\pi}^{2} }{2} [/tex]

[tex] - \frac{c}{b} [/tex]

will translate the graph horizontally to the right by

[tex] \frac{c}{b} [/tex]

units.

+d shifts the graph up by d units.

The new function then becomes:

[tex]y = \cot( \frac{ {\pi}^{2} }{2} (x - \frac{\pi}{4} ) )+1[/tex]

mnunes8347

Answer:

The answer is C: He graphed the function y=cot(2x-pi/4)+1 correctly but it was not the right function to graph. He should have graphed y=cot(2x-pi/2)+1.

Step-by-step explanation:

The reason why it is C is because we want a period of pi/2, which would mean that b must be equal to 2 (if you use the period equation for tan and cot, pi/b, in order for pi/b to be equal to pi/2, b must be 2). The form for a trigonometric function is: y = acotb(x-h)+k. And if you notice, the equation he uses has the b already distributed inside the parenthesis, which means that both x and h were already multiplied. If we divide 2x and pi/4 by two, we get x, but h becomes pi/8, which is not equal to pi/4 as required by the problem. The correct equation would be: y = cot(2x-pi/2)+1 because when you divide out the two from inside the parenthesis, you get: y = cot2(x-pi/4)+1, which is the equation that he should've graphed.

I hope this helped you out!

If you have any further questions don't be afraid to ask.

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