Write a tangent function, h(x), such that the halfway points on one period are (-pi/2, -2) and (pi/2, 4).

Thank you for the opportunity to help you with your question!

the tangent function is tan(x+pi/2)-2

Hi Fatin. Would you please show your work - how you figured out the values of b, h, and k, etc.? Thank you.

ok, please wait for a moment, I'm in the middle of an exam, hope you don't mind :) thanks

assume y=a tan(b(x-h))+ka=amplitudeb=w (omega)h=horizontal shiftk=midline

b=2pi/T=2pi/(4pi)=1/2

a=(4-(-2))/2=6/2=3

thus we have y=3tan(0.5x-0.5h)+k, plug in the two points we get:

-2=3tan(-pi/4-0.5h)+k

4=3tan(pi/4-0.5h)+k

if you solve the system of equations, you will get h=pi/2, k=1

thus the equation is y=3 tan(1/2x-pi/4)+1

sorry I got the answer wrong the first time....please don't be mad....I didn't have enough time, sorry...

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