## Lena772 Group Title (a) find an equation of the tangent line to the graph of at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. one year ago one year ago

1. Lena772

function: y= cos 3x point: (pi/4, -sqrt2/2)

2. ganeshie8

where wer u stuck ?

3. ganeshie8

i saw u doing a similar problem the other day ? :)

4. Decart

take the derivative

5. Lena772

I just get confused when trig is added in

6. Decart

do you know what the derivative is

7. Lena772

dy/dx

8. Decart

-3sin3x

9. Decart

right

10. Lena772

yes

11. ganeshie8

so, slope = m = dy/dx = -3sin3x

12. ganeshie8

since u want the slope at point: (pi/4, -sqrt2/2) evaluate slope at x = pi/4

13. ganeshie8

dy/dx at x = pi/4 :- -3sin(3*pi/4) = ?

14. Lena772

-0.123

15. Lena772

@ganeshie8

16. ganeshie8

that doesnt look right you should get :- dy/dx at x = pi/4 :- -3sin(3*pi/4) = $$\large \frac{-3}{\sqrt{2}}$$

17. ganeshie8

now that u knw slope, m = $$\large \frac{-3}{\sqrt{2}}$$ point: (pi/4, -sqrt2/2) wats the equation of tangent in point slope form ?

18. Lena772

y-(-sqrt2/3)=(-3/sqrt2)(x-(pi/4)

19. Lena772

?

20. Lena772

@hartnn

21. ganeshie8

thats right !

22. Lena772

So that's for a? What would the graph look like for b? Do I just put that equation into Geogebra?

23. Lena772

@ganeshie8

24. ganeshie8

yup graph all 3 below :- 1) function: y= cos 3x 2) point: (pi/4, -sqrt2/2) 3) tangent : y-(-sqrt2/$$\color{red}{2}$$)=(-3/sqrt2)(x-(pi/4)

25. ganeshie8

u should get exactly like the one attached