# Thread: Graphing Trig Expression help

1. ## Graphing Trig Expression help

I need some help finding: the first local minimum point to the right of the y-axis of : y=1/2 sin(6(x-pie/4)+10

this is what i have done
A: 1/2
period: 2pie/6 = pie/3
vertical displacement: 10

max at 10.5 min at 9.5

i know the y is 9.5 and something gone weirdly wrong when i tired to find the x value.

period is pie/3 so divided by 4 to find the distance between two points = pie/12

then X= pie/3 - pie/12
X= pie/6 this is the correct answer on the answer key , i got it right.

further on, i tired to find the next point which is x,10. I minus pie/6 by pie/12..i got pie/6.

Help
how come i got another pie/6?..

Thank You

2. Originally Posted by hovermet
I need some help finding: the first local minimum point to the right of the y-axis of : y=1/2 sin(6(x-pie/4)+10

this is what i have done
A: 1/2
period: 2pie/6 = pie/3
vertical displacement: 10

max at 10.5 min at 9.5

i know the y is 9.5 and something gone weirdly wrong when i tired to find the x value.

period is pie/3 so divided by 4 to find the distance between two points = pie/12

then X= pie/3 - pie/12
X= pie/6 this is the correct answer on the answer key , i got it right.

further on, i tired to find the next point which is x,10. I minus pie/6 by pie/12..i got pie/6.

Help
how come i got another pie/6?..

Thank You
You are correct that $\displaystyle y = 9.5$.

So $\displaystyle 9.5 = \frac{1}{2}\sin{\left[6\left(x - \frac{\pi}{4}\right)\right]} + 10$

$\displaystyle -0.5 = \frac{1}{2}\sin{\left[6\left(x - \frac{\pi}{4}\right)\right]}$

$\displaystyle -1 = \sin{\left[6\left(x - \frac{\pi}{4}\right)\right]}$

$\displaystyle \frac{3\pi}{2} = 6\left(x - \frac{\pi}{4}\right)$

$\displaystyle \frac{\pi}{4} = x - \frac{\pi}{4}$

$\displaystyle x = \frac{\pi}{2}$.
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